Primitive of Power of Hyperbolic Tangent of a x by Square of Hyperbolic Secant of a x

Theorem

$\ds \int \tanh^n a x \sech^2 a x \rd x = \frac {\tanh^{n + 1} a x} {\paren {n + 1} a} + C$

$\ds z$

$=$

$\ds \tanh a x$

$\ds \leadsto \ \ $

$\ds \frac {\d z} {\d x}$

$=$

$\ds a \sech^2 a x$

$\ds \leadsto \ \ $

$\ds \int \tanh^n a x \sech^2 a x \rd x$

$=$

$\ds \int \frac 1 a z^n \rd z$

$\ds $

$=$

$\ds \frac 1 a \frac {z^{n + 1} } {n + 1}$

$\ds $

$=$

$\ds \frac {\tanh^{n + 1} a x} {\paren {n + 1} a} + C$

substituting for $z$ and simplifying

$\blacksquare$