Primitive of Reciprocal of a x + b squared/Proof 2

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Theorem

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


Proof

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

$\ds \int \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$

where $n \ne 1$.

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

$\blacksquare$