Table of Integrals of Irrational Functions

Integrals Containing x\sqrt{x} and a2±b2xa^2 \pm b^2 x

Notation

Let X=a2±b2xX = a^2 \pm b^2 x and define:

Y={arctan⁡(bxa),for X=a2+b2x,12ln⁡∣a+bxa−bx∣,for X=a2−b2x.Y = \begin{cases} \displaystyle \arctan\left( \frac{b\sqrt{x}}{a} \right), & \text{for } X = a^2 + b^2 x, \\[2ex] \displaystyle \frac{1}{2} \ln \left| \frac{a + b\sqrt{x}}{a - b\sqrt{x}} \right|, & \text{for } X = a^2 - b^2 x. \end{cases}

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Note

In formulas with double signs (±\pm or ∓\mp), the upper sign corresponds to X=a2+b2xX = a^2 + b^2 x, and the lower sign to X=a2−b2xX = a^2 - b^2 x.

∫x dxX=±2xb2∓2ab3 Y\int \frac{\sqrt{x} \, dx}{X} = \pm \frac{2\sqrt{x}}{b^2} \mp \frac{2a}{b^3} \, Y

∫x3 dxX=±23⋅x3b2−2a2xb4+2a3b5 Y\int \frac{\sqrt{x^3} \, dx}{X} = \pm \frac{2}{3} \cdot \frac{\sqrt{x^3}}{b^2} - \frac{2a^2\sqrt{x}}{b^4} + \frac{2a^3}{b^5} \, Y

∫x dxX2=±xb2X±1ab3 Y\int \frac{\sqrt{x} \, dx}{X^2} = \pm \frac{\sqrt{x}}{b^2 X} \pm \frac{1}{a b^3} \, Y

∫x3 dxX2=±2x3b2X+3a2xb4X−3a3b5 Y\int \frac{\sqrt{x^3} \, dx}{X^2} = \pm \frac{2\sqrt{x^3}}{b^2 X} + \frac{3a^2\sqrt{x}}{b^4 X} - \frac{3a^3}{b^5} \, Y

∫dxXx=2ab Y\int \frac{dx}{X \sqrt{x}} = \frac{2}{a b} \, Y

∫dxXx3=−2a3x∓2ba3 Y\int \frac{dx}{X \sqrt{x^3}} = -\frac{2}{a^3 \sqrt{x}} \mp \frac{2b}{a^3} \, Y

∫dxX2x=xa2X+ba3 Y\int \frac{dx}{X^2 \sqrt{x}} = \frac{\sqrt{x}}{a^2 X} + \frac{b}{a^3} \, Y

∫dxX3x3=−2a3Xx∓3b2xa4X∓3b3a5 Y\int \frac{dx}{X^3 \sqrt{x^3}} = -\frac{2}{a^3 X \sqrt{x}} \mp \frac{3b^2 \sqrt{x}}{a^4 X} \mp \frac{3b^3}{a^5} \, Y

Other Integrals Containing x\sqrt{x}

∫x dxa4+x2=−12a2ln⁡(x+a2x+a2x−a2x+a2)+1a2arctan⁡(a2xa2−x)\int \frac{\sqrt{x} \, dx}{a^4 + x^2} = -\frac{1}{2a\sqrt{2}} \ln \left( \frac{x + a\sqrt{2x} + a^2}{x - a\sqrt{2x} + a^2} \right) + \frac{1}{a\sqrt{2}} \arctan \left( \frac{a\sqrt{2x}}{a^2 - x} \right)

∫dx(a4+x2)x=12a32ln⁡(x+a2x+a2x−a2x+a2)+1a32arctan⁡(a2xa2−x)\int \frac{dx}{(a^4 + x^2) \sqrt{x}} = \frac{1}{2a^3\sqrt{2}} \ln \left( \frac{x + a\sqrt{2x} + a^2}{x - a\sqrt{2x} + a^2} \right) + \frac{1}{a^3\sqrt{2}} \arctan \left( \frac{a\sqrt{2x}}{a^2 - x} \right)

∫x dxa4−x2=12aln⁡∣a+xa−x∣−1aarctan⁡(xa)\int \frac{\sqrt{x} \, dx}{a^4 - x^2} = \frac{1}{2a} \ln \left| \frac{a + \sqrt{x}}{a - \sqrt{x}} \right| - \frac{1}{a} \arctan \left( \frac{\sqrt{x}}{a} \right)

∫dx(a4−x2)x=12a3ln⁡∣a+xa−x∣+1a3arctan⁡(xa)\int \frac{dx}{(a^4 - x^2) \sqrt{x}} = \frac{1}{2a^3} \ln \left| \frac{a + \sqrt{x}}{a - \sqrt{x}} \right| + \frac{1}{a^3} \arctan \left( \frac{\sqrt{x}}{a} \right)

Integrals Containing ax+b\sqrt{ax + b}

Notation

X=ax+bX = ax + b

∫X dx=23a X3/2\int \sqrt{X} \, dx = \frac{2}{3a} \, X^{3/2}

∫xX dx=2(3ax−2b)15a2 X3/2\int x \sqrt{X} \, dx = \frac{2(3ax - 2b)}{15a^2} \, X^{3/2}

∫x2X dx=2(15a2x2−12abx+8b2)105a3 X3/2\int x^2 \sqrt{X} \, dx = \frac{2(15a^2 x^2 - 12abx + 8b^2)}{105a^3} \, X^{3/2}

∫dxX=2aX\int \frac{dx}{\sqrt{X}} = \frac{2}{a} \sqrt{X}

∫x dxX=2(ax−2b)3a2X\int \frac{x \, dx}{\sqrt{X}} = \frac{2(ax - 2b)}{3a^2} \sqrt{X}

∫x2 dxX=2(3a2x2−4abx+8b2)15a3X\int \frac{x^2 \, dx}{\sqrt{X}} = \frac{2(3a^2 x^2 - 4abx + 8b^2)}{15a^3} \sqrt{X}

∫dxxX={1bln⁡∣X−bX+b∣,b>0,2−barctan⁡X−b,b<0.\int \frac{dx}{x \sqrt{X}} = \begin{cases} \displaystyle \frac{1}{\sqrt{b}} \ln \left| \frac{\sqrt{X} - \sqrt{b}}{\sqrt{X} + \sqrt{b}} \right|, & b > 0, \\[2ex] \displaystyle \frac{2}{\sqrt{-b}} \arctan \sqrt{\frac{X}{-b}}, & b < 0. \end{cases}

∫Xx dx=2X+b∫dxxX(see 127)\int \frac{\sqrt{X}}{x} \, dx = 2 \sqrt{X} + b \int \frac{dx}{x \sqrt{X}} \qquad \text{(see 127)}

∫dxx2X=−Xbx−a2b∫dxxX(see 127)\int \frac{dx}{x^2 \sqrt{X}} = -\frac{\sqrt{X}}{bx} - \frac{a}{2b} \int \frac{dx}{x \sqrt{X}} \qquad \text{(see 127)}

∫Xx2 dx=−Xx+a2∫dxxX(see 127)\int \frac{\sqrt{X}}{x^2} \, dx = -\frac{\sqrt{X}}{x} + \frac{a}{2} \int \frac{dx}{x \sqrt{X}} \qquad \text{(see 127)}

∫dxxnX=−X(n−1)bxn−1−(2n−3)a(2n−2)b∫dxxn−1X,n>1\int \frac{dx}{x^n \sqrt{X}} = -\frac{\sqrt{X}}{(n-1) b x^{n-1}} - \frac{(2n-3)a}{(2n-2)b} \int \frac{dx}{x^{n-1} \sqrt{X}}, \qquad n>1

∫X3/2 dx=25a X5/2\int X^{3/2} \, dx = \frac{2}{5a} \, X^{5/2}

∫xX3/2 dx=235a2(5ax−2b) X5/2\int x X^{3/2} \, dx = \frac{2}{35a^2} (5a x - 2b) \, X^{5/2}

∫x2X3/2 dx=2315a3(35a2x2−20abx+8b2) X5/2\int x^2 X^{3/2} \, dx = \frac{2}{315a^3} \bigl( 35a^2 x^2 - 20ab x + 8b^2 \bigr) \, X^{5/2}

∫X3/2x dx=23X3/2+2bX+b2∫dxxX(see 127)\int \frac{X^{3/2}}{x} \, dx = \frac{2}{3} X^{3/2} + 2b \sqrt{X} + b^2 \int \frac{dx}{x \sqrt{X}} \qquad \text{(see 127)}

∫x dxX3/2=2a2(X+bX)\int \frac{x \, dx}{X^{3/2}} = \frac{2}{a^2} \left( \sqrt{X} + \frac{b}{\sqrt{X}} \right)

∫x2 dxX3/2=2a3(X3/23−2bX−b2X)\int \frac{x^2 \, dx}{X^{3/2}} = \frac{2}{a^3} \left( \frac{X^{3/2}}{3} - 2b \sqrt{X} - \frac{b^2}{\sqrt{X}} \right)

∫dxxX3/2=2bX+1b∫dxxX(see 127)\int \frac{dx}{x X^{3/2}} = \frac{2}{b\sqrt{X}} + \frac{1}{b} \int \frac{dx}{x \sqrt{X}} \qquad \text{(see 127)}

∫dxx2X3/2=−1bxX−3a2b(2bX+1b∫dxxX)\int \frac{dx}{x^2 X^{3/2}} = -\frac{1}{b x \sqrt{X}} - \frac{3a}{2b} \left( \frac{2}{b\sqrt{X}} + \frac{1}{b} \int \frac{dx}{x \sqrt{X}} \right)

∫Xn/2 dx=2a(n+2) X(n+2)/2,n≠−2\int X^{n/2} \, dx = \frac{2}{a(n+2)} \, X^{(n+2)/2}, \qquad n \neq -2

∫xXn/2 dx=2a2(X(n+4)/2n+4−bX(n+2)/2n+2),n≠−2,−4\int x X^{n/2} \, dx = \frac{2}{a^2} \left( \frac{X^{(n+4)/2}}{n+4} - \frac{b X^{(n+2)/2}}{n+2} \right), \qquad n \neq -2, -4

∫x2Xn/2 dx=2a3(X(n+6)/2n+6−2bX(n+4)/2n+4+b2X(n+2)/2n+2),n≠−2,−4,−6\int x^2 X^{n/2} \, dx = \frac{2}{a^3} \left( \frac{X^{(n+6)/2}}{n+6} - \frac{2b X^{(n+4)/2}}{n+4} + \frac{b^2 X^{(n+2)/2}}{n+2} \right), \qquad n \neq -2, -4, -6

∫Xn/2x dx=2nXn/2+b∫X(n−2)/2x dx,n≠0\int \frac{X^{n/2}}{x} \, dx = \frac{2}{n} X^{n/2} + b \int \frac{X^{(n-2)/2}}{x} \, dx, \qquad n \neq 0

∫dxxXn/2=2(2−n)bX(n−2)/2+1b∫dxxX(n−2)/2,n≠2\int \frac{dx}{x X^{n/2}} = \frac{2}{(2-n)b X^{(n-2)/2}} + \frac{1}{b} \int \frac{dx}{x X^{(n-2)/2}}, \qquad n \neq 2

∫dxx2Xn/2=−1bxX(n−2)/2−na2b∫dxxXn/2\int \frac{dx}{x^2 X^{n/2}} = -\frac{1}{b x X^{(n-2)/2}} - \frac{na}{2b} \int \frac{dx}{x X^{n/2}}

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Note:

In the recursive formulas (131, 138, 139, 143-145), the reduction must be applied successively until reaching a known integral (such as 127).

Integrals Containing ax+b\sqrt{ax + b} and fx+g\sqrt{fx + g}

Notation

X=ax+b,Y=fx+g,Δ=bf−ag.X = ax + b, \qquad Y = fx + g, \qquad \Delta = bf - ag.

∫dxXY={2−af arctan⁡−fXaY,if af<0,1af ln⁡∣aY+fX∣,if af>0.\int \frac{dx}{\sqrt{XY}} = \begin{cases} \displaystyle \frac{2}{\sqrt{-af}} \, \arctan \sqrt{\frac{-fX}{aY}}, & \text{if } af < 0, \\[2ex] \displaystyle \frac{1}{\sqrt{af}} \, \ln \left| \sqrt{aY} + \sqrt{fX} \right|, & \text{if } af > 0. \end{cases}

∫x dxXY=XYaf−ag+bf2af∫dxXY(see 146)\int \frac{x \, dx}{\sqrt{XY}} = \frac{\sqrt{XY}}{af} - \frac{ag + bf}{2af} \int \frac{dx}{\sqrt{XY}} \qquad \text{(see 146)}

∫dxX Y3/2=−2XΔY\int \frac{dx}{\sqrt{X} \, Y^{3/2}} = -\frac{2\sqrt{X}}{\Delta \sqrt{Y}}

∫dxYX={2−Δf arctan⁡fX−Δ,if Δf<0,1Δf ln⁡∣fX−ΔffX+Δf∣,if Δf>0.\int \frac{dx}{Y \sqrt{X}} = \begin{cases} \displaystyle \frac{2}{\sqrt{-\Delta f}} \, \arctan \sqrt{ \frac{fX}{-\Delta} }, & \text{if } \Delta f < 0, \\[2ex] \displaystyle \frac{1}{\sqrt{\Delta f}} \, \ln \left| \frac{\sqrt{fX} - \sqrt{\Delta f}}{\sqrt{fX} + \sqrt{\Delta f}} \right|, & \text{if } \Delta f > 0. \end{cases}

∫XY dx=Δ+2aY4afXY−Δ28af∫dxXY(see 146)\int \sqrt{XY} \, dx = \frac{\Delta + 2aY}{4af} \sqrt{XY} - \frac{\Delta^2}{8af} \int \frac{dx}{\sqrt{XY}} \qquad \text{(see 146)}

∫YX dx=1aXY−Δ2a∫dxXY(see 146)\int \sqrt{\frac{Y}{X}} \, dx = \frac{1}{a} \sqrt{XY} - \frac{\Delta}{2a} \int \frac{dx}{\sqrt{XY}} \qquad \text{(see 146)}

∫XY dx=2Xf+Δf∫dxYX(see 149)\int \frac{\sqrt{X}}{Y} \, dx = \frac{2\sqrt{X}}{f} + \frac{\Delta}{f} \int \frac{dx}{Y\sqrt{X}} \qquad \text{(see 149)}

∫Yn dxX=2(2n+1)a(X Yn−nΔ∫Yn−1 dxX)\int \frac{Y^n \, dx}{\sqrt{X}} = \frac{2}{(2n+1)a} \left( \sqrt{X} \, Y^n - n\Delta \int \frac{Y^{n-1} \, dx}{\sqrt{X}} \right)

∫dxX Yn=−1(n−1)Δ(XYn−1+(2n−3)a2∫dxX Yn−1)\int \frac{dx}{\sqrt{X} \, Y^n} = -\frac{1}{(n-1)\Delta} \left( \frac{\sqrt{X}}{Y^{n-1}} + \frac{(2n-3)a}{2} \int \frac{dx}{\sqrt{X} \, Y^{n-1}} \right)

∫X Yn dx=2(2n+3)f(X Yn+1−Δ2∫Yn dxX)(see 153)\int \sqrt{X} \, Y^n \, dx = \frac{2}{(2n+3)f} \left( \sqrt{X} \, Y^{n+1} - \frac{\Delta}{2} \int \frac{Y^n \, dx}{\sqrt{X}} \right) \qquad \text{(see 153)}

∫X dxYn=−1(n−1)f(XYn−1+a2∫dxX Yn−1)\int \frac{\sqrt{X} \, dx}{Y^n} = -\frac{1}{(n-1)f} \left( \frac{\sqrt{X}}{Y^{n-1}} + \frac{a}{2} \int \frac{dx}{\sqrt{X} \, Y^{n-1}} \right)

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Note:

In the recursive formulas (147, 150-156), the reduction must be applied successively until reaching a known integral (such as 146 or 149).

Integrals Containing a2−x2\sqrt{a^2 - x^2}

Notation

X=a2−x2X = a^2 - x^2

∫X dx=12(xX+a2arcsin⁡xa)\int \sqrt{X} \, dx = \frac{1}{2} \left( x\sqrt{X} + a^2 \arcsin\frac{x}{a} \right)

∫xX dx=−13X3/2\int x\sqrt{X} \, dx = -\frac{1}{3} X^{3/2}

∫x2X dx=−x4X3/2+a28(xX+a2arcsin⁡xa)\int x^2 \sqrt{X} \, dx = -\frac{x}{4} X^{3/2} + \frac{a^2}{8} \left( x\sqrt{X} + a^2 \arcsin\frac{x}{a} \right)

∫x3X dx=(X5/25−a2X3/23)\int x^3 \sqrt{X} \, dx = \left( \frac{X^{5/2}}{5} - \frac{a^2 X^{3/2}}{3} \right)

∫Xx dx=X−aln⁡∣a+Xx∣\int \frac{\sqrt{X}}{x} \, dx = \sqrt{X} - a \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫Xx2 dx=−Xx−arcsin⁡xa\int \frac{\sqrt{X}}{x^2} \, dx = -\frac{\sqrt{X}}{x} - \arcsin\frac{x}{a}

∫Xx3 dx=−X2x2+12aln⁡∣a+Xx∣\int \frac{\sqrt{X}}{x^3} \, dx = -\frac{\sqrt{X}}{2x^2} + \frac{1}{2a} \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫dxX=arcsin⁡xa\int \frac{dx}{\sqrt{X}} = \arcsin\frac{x}{a}

∫x dxX=−X\int \frac{x\,dx}{\sqrt{X}} = -\sqrt{X}

∫x2 dxX=−x2X+a22arcsin⁡xa\int \frac{x^2\,dx}{\sqrt{X}} = -\frac{x}{2}\sqrt{X} + \frac{a^2}{2} \arcsin\frac{x}{a}

∫x3 dxX=X3/23−a2X\int \frac{x^3\,dx}{\sqrt{X}} = \frac{X^{3/2}}{3} - a^2 \sqrt{X}

∫dxxX=−1aln⁡∣a+Xx∣\int \frac{dx}{x\sqrt{X}} = -\frac{1}{a} \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫dxx2X=−Xa2x\int \frac{dx}{x^2\sqrt{X}} = -\frac{\sqrt{X}}{a^2 x}

∫dxx3X=−X2a2x2−12a3ln⁡∣a+Xx∣\int \frac{dx}{x^3\sqrt{X}} = -\frac{\sqrt{X}}{2a^2 x^2} - \frac{1}{2a^3} \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫X3/2 dx=14(xX3/2+3a2x2X+3a42arcsin⁡xa)\int X^{3/2} \, dx = \frac{1}{4} \left( x X^{3/2} + \frac{3a^2 x}{2} \sqrt{X} + \frac{3a^4}{2} \arcsin\frac{x}{a} \right)

∫xX3/2 dx=−15X5/2\int x X^{3/2} \, dx = -\frac{1}{5} X^{5/2}

∫x2X3/2 dx=−x6X5/2+a224(xX3/2+3a2x2X+3a42arcsin⁡xa)\int x^2 X^{3/2} \, dx = -\frac{x}{6} X^{5/2} + \frac{a^2}{24} \left( x X^{3/2} + \frac{3a^2 x}{2} \sqrt{X} + \frac{3a^4}{2} \arcsin\frac{x}{a} \right)

∫x3X3/2 dx=X7/27−a2X5/25\int x^3 X^{3/2} \, dx = \frac{X^{7/2}}{7} - \frac{a^2 X^{5/2}}{5}

∫X3/2x dx=X3/23+a2X−a3ln⁡∣a+Xx∣\int \frac{X^{3/2}}{x} \, dx = \frac{X^{3/2}}{3} + a^2 \sqrt{X} - a^3 \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫X3/2x2 dx=−X3/2x−32xX−3a22arcsin⁡xa\int \frac{X^{3/2}}{x^2} \, dx = -\frac{X^{3/2}}{x} - \frac{3}{2} x \sqrt{X} - \frac{3a^2}{2} \arcsin\frac{x}{a}

∫X3/2x3 dx=−X3/22x2−3X2+3a2ln⁡∣a+Xx∣\int \frac{X^{3/2}}{x^3} \, dx = -\frac{X^{3/2}}{2x^2} - \frac{3\sqrt{X}}{2} + \frac{3a}{2} \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫dxX3/2=xa2X\int \frac{dx}{X^{3/2}} = \frac{x}{a^2 \sqrt{X}}

∫x dxX3/2=1X\int \frac{x\,dx}{X^{3/2}} = \frac{1}{\sqrt{X}}

∫x2 dxX3/2=xX−arcsin⁡xa\int \frac{x^2\,dx}{X^{3/2}} = \frac{x}{\sqrt{X}} - \arcsin\frac{x}{a}

∫x3 dxX3/2=X+a2X\int \frac{x^3\,dx}{X^{3/2}} = \sqrt{X} + \frac{a^2}{\sqrt{X}}

∫dxxX3/2=1a2X−1a3ln⁡∣a+Xx∣\int \frac{dx}{x X^{3/2}} = \frac{1}{a^2 \sqrt{X}} - \frac{1}{a^3} \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫dxx2X3/2=−Xa4x+xa4X\int \frac{dx}{x^2 X^{3/2}} = -\frac{\sqrt{X}}{a^4 x} + \frac{x}{a^4 \sqrt{X}}

∫dxx3X3/2=−12a2x2X+32a4X−32a5ln⁡∣a+Xx∣\int \frac{dx}{x^3 X^{3/2}} = -\frac{1}{2a^2 x^2 \sqrt{X}} + \frac{3}{2a^4 \sqrt{X}} - \frac{3}{2a^5} \ln\left| \frac{a + \sqrt{X}}{x} \right|

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Note:

All integrals assume a>0a > 0 and ∣x∣<a|x| < a so that a2−x2\sqrt{a^2 - x^2} is real.

Integrals Containing x2+a2\sqrt{x^2 + a^2}

Notation

X=x2+a2X = x^2 + a^2

∫X dx=12(xX+a2ln⁡(x+X))+C\int \sqrt{X} \, dx = \frac{1}{2} \left( x\sqrt{X} + a^2 \ln\bigl(x + \sqrt{X}\bigr) \right) + C

∫xX dx=13X3/2\int x\sqrt{X} \, dx = \frac{1}{3} X^{3/2}

∫x2X dx=x4X3/2−a28(xX+a2ln⁡(x+X))+C\int x^2 \sqrt{X} \, dx = \frac{x}{4} X^{3/2} - \frac{a^2}{8} \left( x\sqrt{X} + a^2 \ln\bigl(x + \sqrt{X}\bigr) \right) + C

∫x3X dx=X5/25−a2X3/23\int x^3 \sqrt{X} \, dx = \frac{X^{5/2}}{5} - \frac{a^2 X^{3/2}}{3}

∫Xx dx=X−aln⁡∣a+Xx∣\int \frac{\sqrt{X}}{x} \, dx = \sqrt{X} - a \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫Xx2 dx=−Xx+ln⁡(x+X)+C\int \frac{\sqrt{X}}{x^2} \, dx = -\frac{\sqrt{X}}{x} + \ln\bigl(x + \sqrt{X}\bigr) + C

∫Xx3 dx=−X2x2−12aln⁡∣a+Xx∣\int \frac{\sqrt{X}}{x^3} \, dx = -\frac{\sqrt{X}}{2x^2} - \frac{1}{2a} \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫dxX=ln⁡(x+X)+C\int \frac{dx}{\sqrt{X}} = \ln\bigl(x + \sqrt{X}\bigr) + C

∫x dxX=X\int \frac{x\,dx}{\sqrt{X}} = \sqrt{X}

∫x2 dxX=x2X−a22ln⁡(x+X)+C\int \frac{x^2\,dx}{\sqrt{X}} = \frac{x}{2} \sqrt{X} - \frac{a^2}{2} \ln\bigl(x + \sqrt{X}\bigr) + C

∫x3 dxX=X3/23−a2X\int \frac{x^3\,dx}{\sqrt{X}} = \frac{X^{3/2}}{3} - a^2 \sqrt{X}

∫dxxX=−1aln⁡∣a+Xx∣\int \frac{dx}{x\sqrt{X}} = -\frac{1}{a} \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫dxx2X=−Xa2x\int \frac{dx}{x^2\sqrt{X}} = -\frac{\sqrt{X}}{a^2 x}

∫dxx3X=−X2a2x2+12a3ln⁡∣a+Xx∣\int \frac{dx}{x^3\sqrt{X}} = -\frac{\sqrt{X}}{2a^2 x^2} + \frac{1}{2a^3} \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫X3/2 dx=14(xX3/2+3a2x2X+3a42ln⁡(x+X))+C\int X^{3/2} \, dx = \frac{1}{4} \left( x X^{3/2} + \frac{3a^2 x}{2} \sqrt{X} + \frac{3a^4}{2} \ln\bigl(x + \sqrt{X}\bigr) \right) + C

∫xX3/2 dx=15X5/2\int x X^{3/2} \, dx = \frac{1}{5} X^{5/2}

∫x2X3/2 dx=x6X5/2−a224(xX3/2+3a2x2X+3a42ln⁡(x+X))+C\int x^2 X^{3/2} \, dx = \frac{x}{6} X^{5/2} - \frac{a^2}{24} \left( x X^{3/2} + \frac{3a^2 x}{2} \sqrt{X} + \frac{3a^4}{2} \ln\bigl(x + \sqrt{X}\bigr) \right) + C

∫x3X3/2 dx=X7/27−a2X5/25\int x^3 X^{3/2} \, dx = \frac{X^{7/2}}{7} - \frac{a^2 X^{5/2}}{5}

∫X3/2x dx=X3/23+a2X−a3ln⁡∣a+Xx∣\int \frac{X^{3/2}}{x} \, dx = \frac{X^{3/2}}{3} + a^2 \sqrt{X} - a^3 \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫X3/2x2 dx=−X3/2x+32xX+3a22ln⁡(x+X)+C\int \frac{X^{3/2}}{x^2} \, dx = -\frac{X^{3/2}}{x} + \frac{3}{2} x \sqrt{X} + \frac{3a^2}{2} \ln\bigl(x + \sqrt{X}\bigr) + C

∫X3/2x3 dx=−X3/22x2+32X−3a2ln⁡∣a+Xx∣\int \frac{X^{3/2}}{x^3} \, dx = -\frac{X^{3/2}}{2x^2} + \frac{3}{2} \sqrt{X} - \frac{3a}{2} \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫dxX3/2=xa2X\int \frac{dx}{X^{3/2}} = \frac{x}{a^2 \sqrt{X}}

∫x dxX3/2=−1X\int \frac{x\,dx}{X^{3/2}} = -\frac{1}{\sqrt{X}}

∫x2 dxX3/2=−xX+ln⁡(x+X)+C\int \frac{x^2\,dx}{X^{3/2}} = -\frac{x}{\sqrt{X}} + \ln\bigl(x + \sqrt{X}\bigr) + C

∫x3 dxX3/2=X+a2X\int \frac{x^3\,dx}{X^{3/2}} = \sqrt{X} + \frac{a^2}{\sqrt{X}}

∫dxxX3/2=1a2X−1a3ln⁡∣a+Xx∣\int \frac{dx}{x X^{3/2}} = \frac{1}{a^2 \sqrt{X}} - \frac{1}{a^3} \ln\left| \frac{a + \sqrt{X}}{x} \right|

∫dxx2X3/2=−Xa4x−xa4X\int \frac{dx}{x^2 X^{3/2}} = -\frac{\sqrt{X}}{a^4 x} - \frac{x}{a^4 \sqrt{X}}

∫dxx3X3/2=−12a2x2X−32a4X+32a6ln⁡∣a+Xx∣\int \frac{dx}{x^3 X^{3/2}} = -\frac{1}{2a^2 x^2 \sqrt{X}} - \frac{3}{2a^4 \sqrt{X}} + \frac{3}{2a^6} \ln\left| \frac{a + \sqrt{X}}{x} \right|

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Note:

All integrals assume a>0a > 0. The notation Arsh⁡(x/a)\operatorname{Arsh}(x/a) is equivalent to ln⁡(x+x2+a2)\ln(x + \sqrt{x^2+a^2}).

Integrals Containing x2−a2\sqrt{x^2 - a^2}

Notation

X=x2−a2(x>a>0)X = x^2 - a^2 \quad (x > a > 0)

∫X dx=12(xX−a2ln⁡∣x+X∣)+C\int \sqrt{X} \, dx = \frac{1}{2} \left( x\sqrt{X} - a^2 \ln\bigl|x + \sqrt{X}\bigr| \right) + C

∫xX dx=13X3/2\int x\sqrt{X} \, dx = \frac{1}{3} X^{3/2}

∫x2X dx=x4X3/2+a28(xX−a2ln⁡∣x+X∣)+C\int x^2 \sqrt{X} \, dx = \frac{x}{4} X^{3/2} + \frac{a^2}{8} \left( x\sqrt{X} - a^2 \ln\bigl|x + \sqrt{X}\bigr| \right) + C

∫x3X dx=X5/25+a2X3/23\int x^3 \sqrt{X} \, dx = \frac{X^{5/2}}{5} + \frac{a^2 X^{3/2}}{3}

∫Xx dx=X−aarccos⁡a∣x∣\int \frac{\sqrt{X}}{x} \, dx = \sqrt{X} - a \arccos\frac{a}{|x|}

∫Xx2 dx=−Xx+ln⁡∣x+X∣+C\int \frac{\sqrt{X}}{x^2} \, dx = -\frac{\sqrt{X}}{x} + \ln\bigl|x + \sqrt{X}\bigr| + C

∫Xx3 dx=−X2x2+12aarccos⁡a∣x∣\int \frac{\sqrt{X}}{x^3} \, dx = -\frac{\sqrt{X}}{2x^2} + \frac{1}{2a} \arccos\frac{a}{|x|}

∫dxX=ln⁡∣x+X∣+C\int \frac{dx}{\sqrt{X}} = \ln\bigl|x + \sqrt{X}\bigr| + C

∫x dxX=X\int \frac{x\,dx}{\sqrt{X}} = \sqrt{X}

∫x2 dxX=x2X+a22ln⁡∣x+X∣+C\int \frac{x^2\,dx}{\sqrt{X}} = \frac{x}{2} \sqrt{X} + \frac{a^2}{2} \ln\bigl|x + \sqrt{X}\bigr| + C

∫x3 dxX=X3/23+a2X\int \frac{x^3\,dx}{\sqrt{X}} = \frac{X^{3/2}}{3} + a^2 \sqrt{X}

∫dxxX=1aarccos⁡a∣x∣\int \frac{dx}{x\sqrt{X}} = \frac{1}{a} \arccos\frac{a}{|x|}

∫dxx2X=Xa2x\int \frac{dx}{x^2\sqrt{X}} = \frac{\sqrt{X}}{a^2 x}

∫dxx3X=X2a2x2+12a3arccos⁡a∣x∣\int \frac{dx}{x^3\sqrt{X}} = \frac{\sqrt{X}}{2a^2 x^2} + \frac{1}{2a^3} \arccos\frac{a}{|x|}

∫X3/2 dx=14(xX3/2+3a2x2X−3a42ln⁡∣x+X∣)+C\int X^{3/2} \, dx = \frac{1}{4} \left( x X^{3/2} + \frac{3a^2 x}{2} \sqrt{X} - \frac{3a^4}{2} \ln\bigl|x + \sqrt{X}\bigr| \right) + C

∫xX3/2 dx=15X5/2\int x X^{3/2} \, dx = \frac{1}{5} X^{5/2}

∫x2X3/2 dx=x6X5/2+a224(xX3/2+3a2x2X−3a42ln⁡∣x+X∣)+C\int x^2 X^{3/2} \, dx = \frac{x}{6} X^{5/2} + \frac{a^2}{24} \left( x X^{3/2} + \frac{3a^2 x}{2} \sqrt{X} - \frac{3a^4}{2} \ln\bigl|x + \sqrt{X}\bigr| \right) + C

∫x3X3/2 dx=X7/27+a2X5/25\int x^3 X^{3/2} \, dx = \frac{X^{7/2}}{7} + \frac{a^2 X^{5/2}}{5}

∫X3/2x dx=X3/23−a2X+a3arccos⁡a∣x∣\int \frac{X^{3/2}}{x} \, dx = \frac{X^{3/2}}{3} - a^2 \sqrt{X} + a^3 \arccos\frac{a}{|x|}

∫X3/2x2 dx=−X3/2x+32xX+3a22ln⁡∣x+X∣+C\int \frac{X^{3/2}}{x^2} \, dx = -\frac{X^{3/2}}{x} + \frac{3}{2} x \sqrt{X} + \frac{3a^2}{2} \ln\bigl|x + \sqrt{X}\bigr| + C

∫X3/2x3 dx=−X3/22x2−32X−3a2arccos⁡a∣x∣\int \frac{X^{3/2}}{x^3} \, dx = -\frac{X^{3/2}}{2x^2} - \frac{3}{2} \sqrt{X} - \frac{3a}{2} \arccos\frac{a}{|x|}

∫dxX3/2=−xa2X\int \frac{dx}{X^{3/2}} = -\frac{x}{a^2 \sqrt{X}}

∫x dxX3/2=−1X\int \frac{x\,dx}{X^{3/2}} = -\frac{1}{\sqrt{X}}

∫x2 dxX3/2=−xX−ln⁡∣x+X∣+C\int \frac{x^2\,dx}{X^{3/2}} = -\frac{x}{\sqrt{X}} - \ln\bigl|x + \sqrt{X}\bigr| + C

∫x3 dxX3/2=X−a2X\int \frac{x^3\,dx}{X^{3/2}} = \sqrt{X} - \frac{a^2}{\sqrt{X}}

∫dxxX3/2=−1a2X−1a3arccos⁡a∣x∣\int \frac{dx}{x X^{3/2}} = -\frac{1}{a^2 \sqrt{X}} - \frac{1}{a^3} \arccos\frac{a}{|x|}

∫dxx2X3/2=−Xa4x+xa4X\int \frac{dx}{x^2 X^{3/2}} = -\frac{\sqrt{X}}{a^4 x} + \frac{x}{a^4 \sqrt{X}}

∫dxx3X3/2=−12a2x2X+32a4X−32a6arccos⁡a∣x∣\int \frac{dx}{x^3 X^{3/2}} = -\frac{1}{2a^2 x^2 \sqrt{X}} + \frac{3}{2a^4 \sqrt{X}} - \frac{3}{2a^6} \arccos\frac{a}{|x|}

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Note:

All integrals assume x>a>0x > a > 0. For x<−ax < -a, one can use the substitution x=−ux = -u and apply these formulas with absolute values. The function arccos⁡(a/x)\arccos(a/x) can also be expressed as \arcsec(x/a)\arcsec(x/a).

Integrals Containing ax2+bx+c\sqrt{ax^2 + bx + c}

Notation

X=ax2+bx+c,Δ=4ac−b2,k=4aΔ.X = ax^2 + bx + c, \qquad \Delta = 4ac - b^2, \qquad k = \frac{4a}{\Delta}.

∫dxX={1aln⁡∣2aX+2ax+b∣+C,a>0,1−aarcsin⁡2ax+b−Δ+C,a<0, Δ<0.\int \frac{dx}{\sqrt{X}} = \begin{cases} \displaystyle \frac{1}{\sqrt{a}} \ln\left| 2\sqrt{aX} + 2ax + b \right| + C, & a > 0, \\[2ex] \displaystyle \frac{1}{\sqrt{-a}} \arcsin\frac{2ax+b}{\sqrt{-\Delta}} + C, & a < 0,\ \Delta < 0. \end{cases}

∫dxxX=−1cln⁡∣2cX+2c+bxx∣(c>0)\int \frac{dx}{x\sqrt{X}} = -\frac{1}{\sqrt{c}} \ln\left| \frac{2\sqrt{cX} + 2c + bx}{x} \right| \quad (c > 0)

∫dxX3/2=2(2ax+b)ΔX\int \frac{dx}{X^{3/2}} = \frac{2(2ax+b)}{\Delta\sqrt{X}}

∫dxX(2n+1)/2=2(2ax+b)(2n−1)ΔX(2n−1)/2+2k(n−1)2n−1∫dxX(2n−1)/2\int \frac{dx}{X^{(2n+1)/2}} = \frac{2(2ax+b)}{(2n-1)\Delta X^{(2n-1)/2}} + \frac{2k(n-1)}{2n-1} \int \frac{dx}{X^{(2n-1)/2}}

∫X dx=(2ax+b)X4a+Δ8a∫dxX(see 241)\int \sqrt{X} \, dx = \frac{(2ax+b)\sqrt{X}}{4a} + \frac{\Delta}{8a} \int \frac{dx}{\sqrt{X}} \qquad \text{(see 241)}

∫X3/2 dx=(2ax+b)XX8a+3Δ16a∫X dx(see 245)\int X^{3/2} \, dx = \frac{(2ax+b)X\sqrt{X}}{8a} + \frac{3\Delta}{16a} \int \sqrt{X} \, dx \qquad \text{(see 245)}

∫X5/2 dx=(2ax+b)X2X12a+5Δ24a∫X3/2 dx(see 246)\int X^{5/2} \, dx = \frac{(2ax+b)X^2\sqrt{X}}{12a} + \frac{5\Delta}{24a} \int X^{3/2} \, dx \qquad \text{(see 246)}

∫X(2n+1)/2 dx=(2ax+b)X(2n−1)/2X4an+(2n−1)Δ8an∫X(2n−3)/2 dx\int X^{(2n+1)/2} \, dx = \frac{(2ax+b)X^{(2n-1)/2}\sqrt{X}}{4an} + \frac{(2n-1)\Delta}{8an} \int X^{(2n-3)/2} \, dx

∫x dxX=Xa−b2a∫dxX(see 241)\int \frac{x\,dx}{\sqrt{X}} = \frac{\sqrt{X}}{a} - \frac{b}{2a} \int \frac{dx}{\sqrt{X}} \qquad \text{(see 241)}

∫x dxX3/2=−2(bx+2c)ΔX\int \frac{x\,dx}{X^{3/2}} = -\frac{2(bx+2c)}{\Delta\sqrt{X}}

∫x dxX(2n+1)/2=−1(2n−1)aX(2n−1)/2−b2a∫dxX(2n+1)/2(see 244)\int \frac{x\,dx}{X^{(2n+1)/2}} = -\frac{1}{(2n-1)a X^{(2n-1)/2}} - \frac{b}{2a} \int \frac{dx}{X^{(2n+1)/2}} \qquad \text{(see 244)}

∫x2 dxX=(x2a−3b4a2)X+3b2−4ac8a2∫dxX(see 241)\int \frac{x^2\,dx}{\sqrt{X}} = \left( \frac{x}{2a} - \frac{3b}{4a^2} \right) \sqrt{X} + \frac{3b^2 - 4ac}{8a^2} \int \frac{dx}{\sqrt{X}} \qquad \text{(see 241)}

∫x2 dxX3/2=(2b2−4ac)x+2bcaΔX+1a∫dxX(see 241)\int \frac{x^2\,dx}{X^{3/2}} = \frac{(2b^2-4ac)x + 2bc}{a\Delta\sqrt{X}} + \frac{1}{a} \int \frac{dx}{\sqrt{X}} \qquad \text{(see 241)}

∫xX dx=X3/23a−b2a∫X dx(see 245)\int x\sqrt{X} \, dx = \frac{X^{3/2}}{3a} - \frac{b}{2a} \int \sqrt{X} \, dx \qquad \text{(see 245)}

∫xX3/2 dx=X5/25a−b2a∫X3/2 dx(see 246)\int x X^{3/2} \, dx = \frac{X^{5/2}}{5a} - \frac{b}{2a} \int X^{3/2} \, dx \qquad \text{(see 246)}

∫xX(2n+1)/2 dx=X(2n+3)/2(2n+3)a−b2a∫X(2n+1)/2 dx(see 248)\int x X^{(2n+1)/2} \, dx = \frac{X^{(2n+3)/2}}{(2n+3)a} - \frac{b}{2a} \int X^{(2n+1)/2} \, dx \qquad \text{(see 248)}

∫x2X dx=xX3/24a−5b12a⋅X3/2a+5b2−4ac8a2∫X dx(see 245)\int x^2 \sqrt{X} \, dx = \frac{x X^{3/2}}{4a} - \frac{5b}{12a} \cdot \frac{X^{3/2}}{a} + \frac{5b^2 - 4ac}{8a^2} \int \sqrt{X} \, dx \qquad \text{(see 245)}

∫dxxX={−1cln⁡∣2cX+2c+bxx∣,c>0,1−carcsin⁡bx+2c∣x∣−Δ,c<0.\int \frac{dx}{x\sqrt{X}} = \begin{cases} \displaystyle -\frac{1}{\sqrt{c}} \ln\left| \frac{2\sqrt{cX} + 2c + bx}{x} \right|, & c > 0, \\[2ex] \displaystyle \frac{1}{\sqrt{-c}} \arcsin\frac{bx+2c}{|x|\sqrt{-\Delta}}, & c < 0. \end{cases}

∫dxx2X=−Xcx−b2c∫dxxX(see 258)\int \frac{dx}{x^2\sqrt{X}} = -\frac{\sqrt{X}}{cx} - \frac{b}{2c} \int \frac{dx}{x\sqrt{X}} \qquad \text{(see 258)}

∫Xx dx=X+b2∫dxX+c∫dxxX(see 241, 258)\int \frac{\sqrt{X}}{x} \, dx = \sqrt{X} + \frac{b}{2} \int \frac{dx}{\sqrt{X}} + c \int \frac{dx}{x\sqrt{X}} \qquad \text{(see 241, 258)}

∫Xx2 dx=−Xx+a∫dxX+b2∫dxxX(see 241, 258)\int \frac{\sqrt{X}}{x^2} \, dx = -\frac{\sqrt{X}}{x} + a \int \frac{dx}{\sqrt{X}} + \frac{b}{2} \int \frac{dx}{x\sqrt{X}} \qquad \text{(see 241, 258)}

∫X(2n+1)/2x dx=X(2n+1)/22n+1+b2∫X(2n−1)/2 dx+c∫X(2n−1)/2x dx\int \frac{X^{(2n+1)/2}}{x} \, dx = \frac{X^{(2n+1)/2}}{2n+1} + \frac{b}{2} \int X^{(2n-1)/2} \, dx + c \int \frac{X^{(2n-1)/2}}{x} \, dx

∫dxxax4+bx=−2bxax4+bx\int \frac{dx}{x\sqrt{ax^4 + bx}} = -\frac{2}{bx} \sqrt{ax^4 + bx}

∫dx2ax−x2=arcsin⁡x−aa\int \frac{dx}{\sqrt{2ax - x^2}} = \arcsin\frac{x-a}{a}

∫x dx2ax−x2=−2ax−x2+aarcsin⁡x−aa\int \frac{x\,dx}{\sqrt{2ax - x^2}} = -\sqrt{2ax - x^2} + a \arcsin\frac{x-a}{a}

∫2ax−x2 dx=x−a22ax−x2+a22arcsin⁡x−aa\int \sqrt{2ax - x^2} \, dx = \frac{x-a}{2} \sqrt{2ax - x^2} + \frac{a^2}{2} \arcsin\frac{x-a}{a}

∫dx(ax3+b)fx3+g={1bag−bfarctan⁡ag−bffx3+gbfx3+g−gax3+b,ag>bf,12bbf−agln⁡∣bfx3+g+gax3+bbfx3+g−gax3+b∣,ag<bf.\int \frac{dx}{(ax^3 + b)\sqrt{fx^3 + g}} = \begin{cases} \displaystyle \frac{1}{\sqrt{b}\sqrt{ag-bf}} \arctan\frac{\sqrt{ag-bf}\sqrt{fx^3+g}}{\sqrt{b}\sqrt{fx^3+g} - \sqrt{g}\sqrt{ax^3+b}}, & ag>bf, \\[2ex] \displaystyle \frac{1}{2\sqrt{b}\sqrt{bf-ag}} \ln\left| \frac{\sqrt{b}\sqrt{fx^3+g} + \sqrt{g}\sqrt{ax^3+b}}{\sqrt{b}\sqrt{fx^3+g} - \sqrt{g}\sqrt{ax^3+b}} \right|, & ag<bf. \end{cases}

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Note:

The formulas assume the radicals are defined in the integration domain. In recursive formulas, the reduction must be applied successively until reaching a known integral.

Integrals Containing Other Irrational Expressions

∫ax+bn dx=n(n+1)a(ax+b)n+1n+C\int \sqrt[n]{ax + b} \, dx = \frac{n}{(n+1)a} (ax + b)^{\frac{n+1}{n}} + C

∫dxax+bn=n(n−1)a(ax+b)n−1n+C(n≠1)\int \frac{dx}{\sqrt[n]{ax + b}} = \frac{n}{(n-1)a} (ax + b)^{\frac{n-1}{n}} + C \quad (n \neq 1)

∫dxxxn+an=−2naln⁡∣a+xn+anxn/2∣+C\int \frac{dx}{x \sqrt{x^n + a^n}} = -\frac{2}{na} \ln\left| \frac{a + \sqrt{x^n + a^n}}{x^{n/2}} \right| + C

∫dxxxn−an=2naarccos⁡(axn/2)+C(x>a>0)\int \frac{dx}{x \sqrt{x^n - a^n}} = \frac{2}{na} \arccos\left( \frac{a}{x^{n/2}} \right) + C \quad (x > a > 0)

∫x dxan−xn=23arcsin⁡(xn/2an/2)+C\int \frac{\sqrt{x} \, dx}{\sqrt{a^n - x^n}} = \frac{2}{3} \arcsin\left( \frac{x^{n/2}}{a^{n/2}} \right) + C

Reduction Formulas for the Binomial Differential Integral

For the binomial integral ∫xm(axn+b)p dx\displaystyle \int x^m (ax^n + b)^p \, dx:

∫xm(axn+b)p dx=1m+np+1[xm+1(axn+b)p+npb∫xm(axn+b)p−1 dx]\int x^m (ax^n + b)^p \, dx = \frac{1}{m+np+1} \left[ x^{m+1} (ax^n + b)^p + npb \int x^m (ax^n + b)^{p-1} \, dx \right]

=1bn(p+1)[−xm+1(axn+b)p+1+(m+n+np+1)∫xm(axn+b)p+1 dx]= \frac{1}{bn(p+1)} \left[ -x^{m+1} (ax^n + b)^{p+1} + (m+n+np+1) \int x^m (ax^n + b)^{p+1} \, dx \right]

=1(m+1)b[xm+1(axn+b)p+1−a(m+n+np+1)∫xm+n(axn+b)p dx]= \frac{1}{(m+1)b} \left[ x^{m+1} (ax^n + b)^{p+1} - a(m+n+np+1) \int x^{m+n} (ax^n + b)^p \, dx \right]

=1a(m+np+1)[xm−n+1(axn+b)p+1−(m−n+1)b∫xm−n(axn+b)p dx]= \frac{1}{a(m+np+1)} \left[ x^{m-n+1} (ax^n + b)^{p+1} - (m-n+1)b \int x^{m-n} (ax^n + b)^p \, dx \right]

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Note:

The reduction formulas are useful for lowering the exponent pp or the degree mm of the binomial integral. They are applied successively until an elementary integral is obtained.