Primitive of Power of x by Inverse Hyperbolic Tangent of x over a

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Theorem

$\displaystyle \int x^m \tanh^{-1} \frac x a \ \mathrm d x = \frac {x^{m + 1} } {m + 1} \tanh^{-1} \frac x a - \frac a {m + 1} \int \frac {x^{m + 1} } {a^2 - x^2} \ \mathrm d x + C$


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle \tanh^{-1} \frac x a\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle \frac a {a^2 - x^2}\) Derivative of $\tanh^{-1} \dfrac x a$


and let:

\(\displaystyle \frac {\mathrm d v} {\mathrm d x}\) \(=\) \(\displaystyle x^m\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {x^{m + 1} } {m + 1}\) Primitive of Power


Then:

\(\displaystyle \int \frac {\tanh^{-1} \dfrac x a \ \mathrm d x} {x^2}\) \(=\) \(\displaystyle \left({\tanh^{-1} \frac x a}\right) \left({\frac {x^{m + 1} } {m + 1} }\right) - \int \left({\frac {x^{m + 1} } {m + 1} }\right) \left({\frac a {a^2 - x^2} }\right) \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^{m + 1} } {m + 1} \tanh^{-1} \frac x a - \frac a {m + 1} \int \frac {x^{m + 1} } {a^2 - x^2} \ \mathrm d x + C\) simplifying

$\blacksquare$


Also see


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