Primitive of Reciprocal of a x + b squared/Proof 1

Theorem

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

Proof

Let $u = a x + b$.

Then:

\(\ds \int \frac {\d x} {\paren {a x + b}^2}\)

\(=\)

\(\ds \frac 1 a \int \frac {\d u} {u^2}\)

Primitive of Function of $a x + b$

\(\ds \)

\(=\)

\(\ds \frac 1 a \frac {-1} u + C\)

Primitive of Power

\(\ds \)

\(=\)

\(\ds -\frac 1 {a \paren {a x + b} } + C\)

substituting for $u$

$\blacksquare$