Primitive of Power of x by Logarithm of x squared plus a squared

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Theorem

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


Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \map \ln {x^2 + a^2}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac {2 x} {x^2 + a^2}\) Derivative of $\ln x$, Derivative of Power, Chain Rule for Derivatives


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds x^m\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {x^{m + 1} } {m + 1}\) Primitive of Power


Then:

\(\ds \int x^m \map \ln {x^2 + a^2} \rd x\) \(=\) \(\ds \frac {x^{m + 1} } {m + 1} \map \ln {x^2 + a^2} - \int \frac {x^{m + 1} } {m + 1} \frac {2 x \rd x} {x^2 + a^2} + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^{m + 1} \map \ln {x^2 + a^2} } {m + 1} - \frac 2 {m + 1} \int \frac {x^{m + 2} } {x^2 + a^2} \d x\) Primitive of Constant Multiple of Function

$\blacksquare$


Also see


Sources

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