Primitive of Reciprocal of a x squared plus b x plus c/c equal to 0/Proof 1

Theorem

Let $c = 0$.

Then:

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

Proof

First:

\(\ds c\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

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

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \frac 1 b \ln \size {\frac x {a x + b} } + C\)

Primitive of $\dfrac 1 {x \paren {a x + b} }$

$\blacksquare$