# Algebraic Integrals Reference

This is a reference of some algebraic integrals. The actual integral may evaluate to transcendental functions. The integrals are divided by the form of the algebraic expression.

## Form: a + bx

$$\displaystyle \int \dfrac{1}{a+bx}dx = \dfrac{1}{b}\ln(a+bx)+C$$

$$\displaystyle \int \dfrac{x}{a+bx}dx = \dfrac{1}{b^2}\left[bx-a\ln(a+bx)\right]+C$$

$$\displaystyle \int \dfrac{1}{ax+bx^n}dx = \dfrac{1}{a(1-n)}\ln\left(b+\dfrac{a}{x^{n-1}}\right)+C$$

$$\displaystyle \int \dfrac{1}{\sqrt{x}(a^2+b^2x)}dx = \dfrac{2}{ab}\tan^{-1}\left(\dfrac{b}{a}\sqrt{x}\right)+C$$

$$\displaystyle \int \dfrac{1}{\sqrt{x}(a^2-b^2x)}dx = \dfrac{2}{ab}\tanh^{-1}\left(\dfrac{b}{a}\sqrt{x}\right)+C$$

$$\displaystyle \int \dfrac{1}{a+b\sqrt{x}}dx = \dfrac{2\sqrt{x}}{b} - \dfrac{2a}{b^2}\ln(a+b\sqrt{x})+C$$

$$\displaystyle \int \dfrac{1}{\sqrt{a+b\sqrt{x}}}dx = \dfrac{4\sqrt{ax + bx\sqrt{x}}}{3b} - \dfrac{8a\sqrt{a + b \sqrt{x}}}{3b^2} + C$$

## Form: a2 ± x2

$$\displaystyle \int \dfrac{1}{a^2+x^2}dx = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+C = -\dfrac{1}{a}\tan^{-1}\dfrac{a}{x}+C$$

$$\displaystyle \int \dfrac{1}{a^2-x^2}dx = \dfrac{1}{a}\tanh^{-1}\dfrac{x}{a} = \dfrac{1}{a}\tanh^{-1}\dfrac{a}{x}+C$$

$$\displaystyle \int \sqrt{r^2-x^2}\text{ }dx = \dfrac{1}{2}r^2\arcsin\dfrac{x}{r} + \dfrac{1}{2}x\sqrt{r^2-x^2}+C =$$ $$\dfrac{1}{2}r^2\arctan\dfrac{x}{\sqrt{r^2-x^2}} + \dfrac{1}{2}x\sqrt{r^2-x^2}+C$$

$$\displaystyle \int \sqrt{x^2\pm a^2}\text{ }dx = \dfrac{1}{2}a^2\sinh^{-1}\dfrac{x}{a} + \dfrac{1}{2}x\sqrt{x^2\pm a^2}+C =$$ $$\dfrac{1}{2}a^2\ln\left(x+\sqrt{x^2\pm a^2} \right) + \dfrac{1}{2}x\sqrt{x^2\pm a^2}+C$$

$$\displaystyle \int \dfrac{1}{\sqrt{a^2-x^2}}dx = \arcsin\dfrac{x}{a}+C$$

$$\displaystyle \int \dfrac{1}{\sqrt{x^2 \pm a^2}}dx = \ln\left( x + \sqrt{x^2\pm a^2} \right)+C$$

$$\displaystyle \int \dfrac{1}{x\sqrt{x^2 - a^2}}dx = -\dfrac{1}{a}\arcsin\dfrac{a}{x}+C$$

$$\displaystyle \int \dfrac{1}{x\sqrt{a^2 - x^2}}dx = -\dfrac{1}{a}\ln\dfrac{a+\sqrt{a^2-x^2}}{x}+C$$

$$\displaystyle \int \dfrac{1}{x\sqrt{x^2 + a^2}}dx = -\dfrac{1}{a}\ln\dfrac{a+\sqrt{x^2+a^2}}{x}+C$$

$$\displaystyle \int \dfrac{1}{\sqrt{ax - x^2}}dx = 2\arcsin\sqrt{\dfrac{x}{a}}+C$$

$$\displaystyle \int \sqrt{\dfrac{1}{x^2}+1}\text{ }dx = \sqrt{1+x^2}-\sinh^{-1}\dfrac{1}{x}+C$$

$$\displaystyle \int \sqrt{\dfrac{1}{x^2}-1}\text{ }dx = \sqrt{1-x^2}-\text{asech}^{-1}x+C$$

$$\displaystyle \int \sqrt{\dfrac{b^2}{x}-a^2}\text{ }dx = \sqrt{b^2x-(ax)^2}+\dfrac{b^2}{a}\sin^{-1}\dfrac{a}{b}\sqrt{x}+C$$

$$\displaystyle \int \sqrt{\dfrac{b^2}{x}+a^2}\text{ }dx = \sqrt{b^2x+(ax)^2}+\dfrac{b^2}{a}\sinh^{-1}\dfrac{a}{b}\sqrt{x}+C$$

## Form: wm ± wm+n

$$\displaystyle \int \dfrac{1}{\sqrt{w^2 \pm w^3}}dw = -2\tanh^{-1}\sqrt{1 \pm w}+C$$

$$\displaystyle \int \dfrac{1}{\sqrt{w^3 \pm w^4}}dw = -2\sqrt{\dfrac{1}{w} \pm 1}+C$$

## Form: $$\dfrac{1}{a^n \pm w^n}$$

$$\displaystyle \int \dfrac{1}{1 \pm w}dw = \pm\ln(1 \pm w) + C$$

$$\displaystyle \int \dfrac{1}{1 + w^2}dw = \arctan w + C$$

$$\displaystyle \int \dfrac{1}{1 - w^2}dw = \tanh^{-1} w + C$$

$$\displaystyle \int \dfrac{1}{1 \pm w^3}dw = \pm\dfrac{1}{3}\ln(1\pm w) \mp \dfrac{1}{6}\ln(1 \mp w + w^2) +$$ $$\dfrac{1}{\sqrt{3}}\arctan\left[ \dfrac{2}{\sqrt{3}}\left(w\mp\dfrac{1}{2}\right) \right] + C$$

$$\displaystyle \int \dfrac{1}{1 + w^4}dw = \dfrac{\sqrt{2}}{4}\tan^{-1}\left( \dfrac{\sqrt{2}w}{1-w^2} \right) +$$$$\dfrac{\sqrt{2}}{4}\tanh^{-1}\left( \dfrac{\sqrt{2}w}{1+w^2} \right) + C$$

$$\displaystyle \int \dfrac{1}{1 - w^4}dw = \dfrac{1}{2}(\tan^{-1}w + \tanh^{-1}w) + C$$

$$\displaystyle \int \dfrac{1}{1 \pm w^6}dw = \dfrac{1}{3}\tan^{-1}w + \dfrac{1}{6}\tan^{-1}\left(\dfrac{w}{1 \mp w^2} \right) +$$$$\dfrac{1}{2\sqrt{3}}\tanh^{-1}\left(\dfrac{\sqrt{3}w}{1\pm w^2} \right) + C$$

$$\displaystyle \int \dfrac{x^2}{a^2 - b^2 x^4}dx = \dfrac{1}{2b\sqrt{ab}}\left[\tanh^{-1}\left( \sqrt{\dfrac{b}{a}}x\right) - \tan^{-1}\left(\sqrt{\dfrac{b}{a}}x \right) \right] + C$$

$$\displaystyle \int \dfrac{x^2}{1 + x^4}dx = \dfrac{\sqrt{2}}{8}\ln\left[ \dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1} \right] +$$$$\dfrac{\sqrt{2}}{4}\tan^{-1}(\sqrt{2}x+1)+$$ $$\dfrac{\sqrt{2}}{4}\tan^{-1}(\sqrt{2}x-1) + C$$

## Form: $$\dfrac{1}{1 \pm y^m \pm y^n}$$

$$\displaystyle \int \dfrac{1}{1 \pm y + y^2}dy =\dfrac{2}{\sqrt{3}}\arctan\left[\dfrac{2}{\sqrt{3}}\left(y \pm \dfrac{1}{2} \right) \right] + C$$

$$\displaystyle \int \dfrac{2-y}{1 - y + y^2}dy =\sqrt{3}\arctan\left[\dfrac{2}{\sqrt{3}}\left(y - \dfrac{1}{2} \right) \right] - \dfrac{1}{2}\ln(1-y+y^2) + C$$

$$\displaystyle \int \dfrac{1}{1 + y^2 + y^4}dy =\dfrac{1}{2\sqrt{3}}\arctan\left(\dfrac{\sqrt{3}y}{1-y^2}\right) + \dfrac{1}{4}\ln\dfrac{1+y+y^2}{1-y+y^2} + C$$

$$\displaystyle \int \dfrac{1}{1 - y^2 + y^4}dy =\dfrac{1}{2\sqrt{3}}\tanh^{-1}\left(\dfrac{\sqrt{3}y}{1+y^2}\right) + \dfrac{1}{2}\tan^{-1}\dfrac{y}{1-y^2} + C$$

## Form: $$(ax^2+bx+c)^n$$

$$\displaystyle \int (ax^2+bx+c)^n \text{ } dx = x(ax^2+bx+c) - n\displaystyle \int x(2ax+b)(ax^2+bx+c)^{n-1}\text{ } dx$$

$$\displaystyle \int \dfrac{1}{ax^2+bx+c}dx = \dfrac{2}{\sqrt{-b^2+4ac}}\tan^{-1}\dfrac{2ax+b}{\sqrt{-b^2+4ac}} + C$$, ($$\color{DarkGreen}-b^2+4ac > 0$$)

$$\displaystyle \int \dfrac{1}{ax^2+bx+c}dx = -\dfrac{2}{\sqrt{b^2-4ac}}\tanh^{-1}\dfrac{2ax+b}{\sqrt{b^2-4ac}} + C$$, ($$\color{DarkGreen}b^2-4ac > 0$$)

$$\displaystyle \int \dfrac{1}{\sqrt{ax^2+bx+c}}dx = \dfrac{1}{\sqrt{a}}\ln\left[\dfrac{2ax+b}{\sqrt{a}}+2\sqrt{ax^2+bx+c} \right] + C$$