# Primitive of Reciprocal of x by a x + b squared

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## Theorem

$\displaystyle \int \frac {\d x} {x \paren {a x + b}^2} = \frac 1 {b \paren {a x + b} } + \frac 1 {b^2} \ln \size {\frac x {a x + b} } + C$

## Proof 1

 $\ds \int \frac {\mathrm d x} {x \left({a x + b}\right)^2}$ $=$ $\ds \int \left({\frac 1 {b^2 x} - \frac a {b^2 \left({a x + b}\right)} - \frac a {b \left({a x + b}\right)^2} }\right) \ \mathrm d x$ Partial Fraction Expansion $\ds$ $=$ $\ds \frac 1 {b^2} \int \frac {\mathrm d x} x - \frac a {b^2} \int \frac {\mathrm d x} {a x + b} - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^2}$ Linear Combination of Integrals $\ds$ $=$ $\ds \frac 1 {b^2} \ln \left\vert{x}\right\vert - \frac a {b^2} \int \frac {\mathrm d x} {a x + b} - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C$ Primitive of Reciprocal $\ds$ $=$ $\ds \frac 1 {b^2} \ln \left\vert{x}\right\vert - \frac a {b^2} \ln \left\vert{a x + b}\right\vert - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C$ Primitive of $\dfrac 1 {a x + b}$ $\ds$ $=$ $\ds \frac 1 {b^2} \ln \left\vert{x}\right\vert - \frac a {b^2} \ln \left\vert{a x + b}\right\vert - \frac a b \frac {-1} {a \left({a x + b}\right)} + C$ Primitive of $\dfrac 1 {\left({a x + b}\right)^2}$ $\ds$ $=$ $\ds \frac 1 {b \left({a x + b}\right)} + \frac 1 {b^2} \ln \left\vert{\frac x {a x + b} }\right\vert + C$ Difference of Logarithms and rearranging

$\blacksquare$

## Proof 2

 $\ds \int \frac {\d x} {x \paren {a x + b}^2}$ $=$ $\ds \int \frac {b \rd x} {b x \paren {a x + b}^2}$ multiplying top and bottom by $b$ $\ds$ $=$ $\ds \int \frac {\paren {a x + b - a x} \rd x} {b x \paren {a x + b}^2}$ adding and subtracting $a x$ $\ds$ $=$ $\ds \frac 1 b \int \frac {\paren {a x + b} \rd x} {x \paren {a x + b}^2} - \frac a b \int \frac {x \rd x} {x \paren {a x + b}^2}$ Linear Combination of Integrals $\ds$ $=$ $\ds \frac 1 b \int \frac {\d x} {x \paren {a x + b} } - \frac a b \int \frac {\d x} {\paren {a x + b}^2}$ simplifying $\ds$ $=$ $\ds \frac 1 b \paren {\frac 1 b \ln \size {\frac x {a x + b} } } - \frac a b \int \frac {\d x} {\paren {a x + b}^2} + C$ Primitive of $\dfrac 1 {x \paren {a x + b} }$ $\ds$ $=$ $\ds \frac 1 {b^2} \ln \size {\frac x {a x + b} } - \frac a b \paren {-\frac 1 {a \paren {a x + b} } } + C$ Primitive of $\dfrac 1 {\paren {a x + b}^2}$ $\ds$ $=$ $\ds \frac 1 {b \paren {a x + b} } + \frac 1 {b^2} \ln \size {\frac x {a x + b} } + C$ simplifying

$\blacksquare$