Primitive of Reciprocal of a x squared + b by Root of c x squared + d/Lemma

Theorem

Let $a, b, c, d \in \R$ be real numbers such that $a d \ne b c$.

Then:

$\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} } = \dfrac {\sqrt c} {2 a} \int \dfrac {\d u} {\paren {\sqrt {\paren {u - \frac d 2}^2 - \frac d 4} } \paren {u - \frac {a d + b c} a} }$

where $u := c x^2 + d$.

Proof

\(\ds u\)

\(=\)

\(\ds c x^2 + d\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds \sqrt {\dfrac {u - d} c}\)

\(\, \ds \text {and} \, \)

\(\ds \frac {\d u} {\d x}\)

\(=\)

\(\ds 2 c x\)

Primitive of Power

Hence:

\(\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} }\)

\(=\)

\(\ds \int \dfrac {\d u} {2 c x \paren {a x^2 + b} \sqrt {c x^2 + d} }\)

Integration by Substitution

\(\ds \)

\(=\)

\(\ds \int \dfrac {\d u} {2 c \sqrt {\frac {u - d} c} \paren {a \paren {\frac {u - d} c} + b} \sqrt u}\)

substituting for $x$

\(\ds \)

\(=\)

\(\ds \int \dfrac {c \sqrt c \d u} {2 c \sqrt {u - d} \paren {a u - a d + b c} \sqrt u}\)

simplifying

\(\ds \)

\(=\)

\(\ds \dfrac {\sqrt c} {2 a} \int \dfrac {\d u} {\paren {\sqrt {\paren {u - \frac d 2}^2 - \frac d 4} } \paren {u - \frac {a d - b c} a} }\)

completing the square

$\blacksquare$