Primitive of x over x squared plus a squared/Corollary

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Theorem

$\ds \int \frac {x \rd x} {a^2 + b^2 x^2} = \frac 1 {2 b^2} \map \ln {a^2 + b^2 x^2} + C$

where $a$ is a non-zero constant.


Proof

Let $z = b x$.

Then:

$\dfrac {\d x} {\d z} = \dfrac 1 b$

Hence:

\(\ds \int \frac {x \rd x} {a^2 + b^2 x^2}\) \(=\) \(\ds \int \dfrac 1 b \frac {\paren {z / b} \d z} {a^2 + z^2}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \dfrac 1 {b^2} \int \frac {z \d z} {a^2 + z^2}\) simplifying, and Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \dfrac 1 {b^2} \cdot \frac 1 2 \map \ln {a^2 + z^2} + C\) Primitive of $\dfrac z {a^2 + z^2}$
\(\ds \) \(=\) \(\ds \dfrac 1 {2 b^2} \map \ln {a^2 + b^2 x^2} + C\) subtituting for $z$ and simplifying

$\blacksquare$


Sources

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