Primitive of a x + b over p x + q

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Theorem

$\displaystyle \int \frac {a x + b} {p x + q} \ \mathrm d x = \frac {a x} p + \frac {b p - a q} {p^2} \ln \left\vert{p x + q}\right\vert + C$


Proof

\(\displaystyle \int \frac {a x + b} {p x + q} \ \mathrm d x\) \(=\) \(\displaystyle a \int \frac {x \ \mathrm d x} {p x + q} + b \int \frac {\mathrm d x} {p x + q}\) Linear Combination of Integrals
\(\displaystyle \) \(=\) \(\displaystyle a \left({\frac x p - \frac q {p^2} \ln \left\vert{p x + q}\right\vert}\right) + b \int \frac {\mathrm d x} {p x + q} + C\) Primitive of $x$ over $a x + b$
\(\displaystyle \) \(=\) \(\displaystyle \frac {a x} p - \frac {a q} {p^2} \ln \left\vert{p x + q}\right\vert + b \left({\frac 1 p \ln \left\vert{p x + q}\right\vert}\right) + C\) Primitive of Reciprocal of $a x + b$
\(\displaystyle \) \(=\) \(\displaystyle \frac {a x} p - \frac {a q} {p^2} \ln \left\vert{p x + q}\right\vert + \frac b p \ln \left\vert{p x + q}\right\vert + C\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {a x} p + \frac {b p - a q} {p^2} \ln \left\vert{p x + q}\right\vert + C\) common denominator

$\blacksquare$


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