Primitive of Inverse Hyperbolic Tangent Function/Proof 1

Theorem

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

for $x^2 < 1$.

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:

$\ds u$

$=$

$\ds \artanh x$

$\ds \leadsto \ \ $

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

$=$

$\ds \frac 1 {1 - x^2}$

and let:

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

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds v$

$=$

$\ds x$

Then:

$\ds \int \artanh x \rd x$

$=$

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

$\ds $

$=$

$\ds x \artanh x - \paren {-\frac 1 2 \map \ln {1 - x^2} } + C$

$\ds $

$=$

$\ds x \artanh x + \frac {\map \ln {1 - x^2} } 2 + C$

simplifying

$\blacksquare$