Primitive of x over x squared plus a squared

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Theorem

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

where $a$ is a non-zero constant.


Corollary

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


Proof 1

\(\ds u\) \(=\) \(\ds x^2 + a^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives and Derivative of Constant
\(\ds \leadsto \ \ \) \(\ds \int \frac {x \rd x} {x^2 + a^2}\) \(=\) \(\ds \frac 1 2 \ln \size {x^2 + a^2} + C\) Primitive of Function under its Derivative
\(\ds \) \(=\) \(\ds \frac 1 2 \, \map \ln {x^2 + a^2} + C\) Absolute Value of Even Power‎

$\blacksquare$


Proof 2

From Primitive of Power of x less one over Power of x plus Power of a:

$\ds \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$


So:

\(\ds \int \frac {x \rd x} {x^2 + a^2}\) \(=\) \(\ds \frac 1 2 \ln \size {x^2 + a^2} + C\) Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$ with $n = 2$
\(\ds \) \(=\) \(\ds \frac 1 2 \, \map \ln {x^2 + a^2} + C\) Absolute Value of Even Power

$\blacksquare$


Sources

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