Primitive of Reciprocal of a x + b squared

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Theorem

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


Proof 1

Let $u = a x + b$.

Then:

\(\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^2}\) \(=\) \(\displaystyle \frac 1 a \int \frac {\mathrm d u} {u^2}\) Primitive of Function of $a x + b$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \frac {-1} u + C\) Primitive of Power
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 {a \left({a x + b}\right)} + C\) substituting for $u$

$\blacksquare$


Proof 2

From Primitive of Power of $a x + b$:

$\displaystyle \int \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a} + C$

where $n \ne 1$.

The result follows by setting $n = -2$.

$\blacksquare$


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