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Integrals involving $ax + b$

\int (a x + b) n d x = ( a x + b ) n + 1 a ( n + 1 ), (for n \neq 1)

$\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)}, \quad (\text{for } n \ne 1)$

\int 1 a x + b d x = 1 a ln | a x + b |

$\int \frac{1}{ax+b}dx = \frac{1}{a}\ln|ax+b|$

\int x (a x + b) n d x = a ( n + 1 ) x - b a 2 ( n + 1 ) ( n + 2 ) (a x + b) n + 1, (for n \neq - 1, n \neq - 2)

$\int x (ax+b)^ndx = \frac{a(n+1)x-b}{a^2(n+1)(n+2)}(ax+b)^{n+1}, \quad (\text{for } n \ne -1, n\ne-2)$

\int x a x + b d x = x 2 - b a 2 ln | a x + b |

$\int \frac{x}{ax+b}dx = \frac{x}{2} - \frac{b}{a^2}\ln|ax+b|$

\int x ( a x + b ) 2 d x = b a 2 ( a x + b ) - 1 a 2 ln | a x + b |

$\int \frac{x}{(ax+b)^2}dx = \frac{b}{a^2(ax+b)} - \frac{1}{a^2}\ln|ax+b|$

\int x 2 a x + b d x = 1 a 3 (( a x + b ) 2 2 - 2 b (a x + b) + b 2 ln | a x + b |)

$\int \frac{x^2}{ax+b} dx = \frac{1}{a^3} \left(\frac{(ax+b)^2}{2}-2b(ax+b)+b^2\ln|ax+b| \right)$

\int x 2 ( a x + b ) 2 d x = 1 a 3 (a x + b - 2 b ln | a x + b | - b 2 a x + b)

$\int \frac{x^2}{(ax+b)^2} dx = \frac{1}{a^3} \left(ax+b-2\,b\,\ln|ax+b| - \frac{b^2}{ax+b}\right)$

\int x 2 ( a x + b ) 3 d x = 1 a 3 (ln | a x + b | + 2 b a x + b - b 2 2 ( a x + b ) 2)

$\int \frac{x^2}{(ax+b)^3} dx = \frac{1}{a^3} \left( \ln|ax+b| + \frac{2b}{ax+b} - \frac{b^2}{2(ax+b)^2} \right)$

\int x 2 ( a x + b ) n d x = 1 a 3 (- ( a x + b ) 3 - n n - 3 + 2 b ( a + b ) 2 - n n - 2 - b 2 ( a x + b ) 1 - n n - 1)

$\int \frac{x^2}{(ax+b)^n} dx = \frac{1}{a^3} \left( -\frac{(ax+b)^{3-n}}{n-3} + \frac{2b(a+b)^{2-n}}{n-2} - \frac{b^2(ax+b)^{1-n}}{n-1}\right)$

\int 1 x ( a x + b ) d x = - 1 b ln ∣ ∣ ∣ a x + b x ∣ ∣ ∣

$\int \frac{1}{x(ax+b)}dx = -\frac{1}{b}\ln\left|\frac{ax+b}{x}\right|$

\int 1 x 2 ( a x + b ) 2 d x = - 1 b x + a b 2 ln ∣ ∣ ∣ a x + b x ∣ ∣ ∣

$\int \frac{1}{x^2(ax+b)^2}dx = -\frac{1}{bx} + \frac{a}{b^2} \ln\left|\frac{ax+b}{x}\right|$

\int 1 x 2 ( a x + b ) 2 d x = - a (1 b 2 ( a x + b ) + 1 a b 2 x - 2 b 3 ln ∣ ∣ ∣ a x + b x ∣ ∣ ∣)

$\int \frac{1}{x^2(ax+b)^2}dx = -a\left(\frac{1}{b^2(ax+b)} + \frac{1}{ab^2x} - \frac{2}{b^3}\ln\left|\frac{ax+b}{x} \right|\right)$

Integrals involving $ax^2 + bx + c$

1 x 2 + a 2 d x = 1 a arctan x a

$\frac{1}{x^2+a^2} dx = \frac{1}{a}\arctan\frac{x}{a}$

1 x 2 - a 2 d x = 1 2 a ln ∣ ∣ ∣ x - a x + a ∣ ∣ ∣

$\frac{1}{x^2 - a^2} dx = \frac{1}{2a} \ln\left| \frac{x-a}{x+a}\right|$

\int 1 a x 2 + b x + c d x = ⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 4 a c - b 2 - - - - - - - \sqrt arctan 2 a x + b 4 a c - b 2 - - - - - - - \sqrt for 4 a c - b 2 > 0 2 b 2 - 4 a c - - - - - - - \sqrt ln ∣ ∣ ∣ 2 a x + b - b 2 - 4 a c - - - - - - - \sqrt 2 a x + b + b 2 - 4 a c - - - - - - - \sqrt ∣ ∣ ∣ for 4 a c - b 2 < 0 - 2 2 a x + b for 4 a c - b 2 = 0

$\int \frac{1}{ax^2 + bx + c} dx = \left\{ \begin{aligned} & \frac{2}{\sqrt{4ac - b^2}} \arctan \frac{2ax + b}{\sqrt{4ac-b^2}} \quad \text{for } 4ac - b^2 > 0 \\ & \frac{2}{\sqrt{b^2 - 4ac}} \ln \left| \frac{2ax + b - \sqrt{b^2-4ac}}{2ax + b + \sqrt{b^2-4ac}} \right| \quad \text{for } 4ac - b^2 < 0 \\ & -\frac{2}{2ax+b} \quad \text{for } 4ac - b^2 = 0 \end{aligned} \right.$

\int x a x 2 + b x + c d x = 1 2 a ln ∣ ∣ a x 2 + b x + c ∣ ∣ - b 2 a \int d x a x 2 + b x + c

$\int \frac{x}{ax^2 + bx +c} dx = \frac{1}{2a} \ln\left|ax^2+bx+c\right| - \frac{b}{2a}\int \frac{dx}{ax^2+bx+c}$

\int 1 ( a x 2 + b x + c ) n d x = 2 a x + b ( n - 1 ) ( 4 a c - b 2 ) ( a x + b x + c ) n - 1 + 2 ( 2 n - 3 ) a ( n - 1 ) ( 4 a c - b 2 ) \int d x ( a x 2 + b x + c ) n - 1

$\int \frac{1}{(ax^2+bx+c)^n} dx = \frac{2ax+b}{(n-1)(4ac-b^2)(ax+bx+c)^{n-1}}+ \frac{2(2n-3)a}{(n-1)(4ac - b^2)}\int\frac{dx}{(ax^2+bx+c)^{n-1}}$

\int 1 x ( a x 2 + b x + c ) d x = 1 2 c ln ∣ ∣ ∣ x 2 a x 2 + b x + c ∣ ∣ ∣ - b 2 c \int 1 a x 2 + b x + c d x

$\int \frac{1}{x(ax^2 + bx + c)} dx = \frac{1}{2c}\ln\left|\frac{x^2}{ax^2+bx+c}\right| - \frac{b}{2c} \int \frac{1}{ax^2+bx+c}dx$

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