Primitive of x squared by Inverse Hyperbolic Cotangent of x over a

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Theorem

$\ds \int x^2 \arcoth \frac x a \rd x = \frac {a x^2} 6 + \frac {x^3} 3 \arcoth \frac x a + \frac {a^3} 6 \map \ln {x^2 - a^2} + C$


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:

\(\ds u\) \(=\) \(\ds \arcoth \frac x a\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac {-a} {x^2 - a^2}\) Derivative of $\arcoth \dfrac x a$


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {x^3} 3\) Primitive of Power


Then:

\(\ds \int x^2 \arcoth \frac x a \rd x\) \(=\) \(\ds \frac {x^3} 3 \arcoth \frac x a - \int \frac {x^3} 3 \paren {\frac {-a} {x^2 - a^2} } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {x^3} 3 \arcoth \frac x a + \frac a 3 \int \frac {x^3 \rd x} {x^2 - a^2} + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {x^3} 3 \arcoth \frac x a + \frac a 3 \paren {\frac {x^2} 2 + \frac {a^2} 2 \map \ln {x^2 - a^2} } + C\) Primitive of $\dfrac {x^3} {x^2 - a^2}$
\(\ds \) \(=\) \(\ds \frac {a x^2} 6 + \frac {x^3} 3 \arcoth \frac x a + \frac {a^3} 6 \map \ln {x^2 - a^2} + C\) simplifying

$\blacksquare$


Also see


Sources

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