Primitive of Reciprocal of x by Root of a x + b

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Theorem

For $a > 0$ and for $x \ne 0$:

$\ds \int \frac {\d x} {x \sqrt {a x + b} } = \begin {cases} \dfrac 1 {\sqrt b} \ln \size {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C & : b > 0 \\ \dfrac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C & : b < 0 \end {cases}$

where $a x + b > 0$.


Proof

Let:

\(\ds u\) \(=\) \(\ds \sqrt {a x + b}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac {u^2 - b} a\)


Thus:

\(\ds \map F {\sqrt {a x + b} }\) \(=\) \(\ds \frac 1 {x \sqrt {a x + b} }\)
\(\ds \leadsto \ \ \) \(\ds \map F u\) \(=\) \(\ds \paren {\frac a {u^2 - b} } \frac 1 u\)


Then:

\(\ds \int \frac {\d x} {x \sqrt {a x + b} }\) \(=\) \(\ds \frac 2 a \int u \paren {\frac a {u^2 - b} } \frac 1 u \rd u\) Primitive of Function of $\sqrt {a x + b}$
\(\ds \) \(=\) \(\ds 2 \int \frac {\d u} {u^2 - b}\) Primitive of Constant Multiple of Function


Let $b > 0$.

Let $d = \sqrt b$.

Then:

\(\ds 2 \int \frac {\d u} {u^2 - b}\) \(=\) \(\ds 2 \int \frac {\d u} {u^2 - d^2}\)
\(\ds \) \(=\) \(\ds 2 \paren {\frac 1 {2 d} } \ln \size {\frac {u - d} {u + d} } + C\) Primitive of Reciprocal of $x^2 - a^2$: Logarithm Form
\(\ds \) \(=\) \(\ds \dfrac 1 {\sqrt b} \ln \size {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C\) substituting for $u$ and $d$

$\Box$


Let $b < 0$.

Let $d = \sqrt {-b}$.

Then:

\(\ds 2 \int \frac {\d u} {u^2 - b}\) \(=\) \(\ds 2 \int \frac {\d u} {u^2 - \paren {-d^2} }\)
\(\ds \) \(=\) \(\ds 2 \int \frac {\d u} {u^2 + d^2}\)
\(\ds \) \(=\) \(\ds 2 \paren {\frac 1 d} \arctan {\frac u d} + C\) Primitive of Reciprocal of $x^2 + a^2$
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C\) substituting for $u$ and $d$

$\blacksquare$


Sources

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