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

Theorem

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

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:

\(\ds u\)

\(=\)

\(\ds \artanh \frac x a\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d u} {\d x}\)

\(=\)

\(\ds \frac a {a^2 - x^2}\)

Derivative of $\artanh \dfrac x a$

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:

\(\ds \int \frac {\artanh \dfrac x a \rd x} {x^2}\)

\(=\)

\(\ds \paren {\artanh \frac x a} \paren {\frac {x^{m + 1} } {m + 1} } - \int \paren {\frac {x^{m + 1} } {m + 1} } \paren {\frac a {a^2 - x^2} } \rd x + C\)

Integration by Parts

\(\ds \)

\(=\)

\(\ds \frac {x^{m + 1} } {m + 1} \artanh \frac x a - \frac a {m + 1} \int \frac {x^{m + 1} } {a^2 - x^2} \rd x + C\)

simplifying

$\blacksquare$

Also see

• Primitive of $x^m \arsinh \dfrac x a$

• Primitive of $x^m \arcosh \dfrac x a$

• Primitive of $x^m \arcoth \dfrac x a$

• Primitive of $x^m \arsech \dfrac x a$

• Primitive of $x^m \arcsch \dfrac x a$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving Inverse Hyperbolic Functions: $14.674$