Primitive of Inverse Hyperbolic Tangent of x over a

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Theorem

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


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:

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


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds x\) Primitive of Constant


Then:

\(\ds \int \artanh \frac x a \rd x\) \(=\) \(\ds x \artanh \frac x a - \int x \paren {\frac a {a^2 - x^2} } \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds x \artanh \frac x a - a \int \frac {x \rd x} {a^2 - x^2} + C\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds x \artanh \frac x a - a \paren {-\frac 1 2 \map \ln {a^2 - x^2} } + C\) Primitive of $\dfrac x {a^2 - x^2}$
\(\ds \) \(=\) \(\ds x \artanh \dfrac x a + \frac {a \map \ln {a^2 - x^2} } 2 + C\) simplifying

$\blacksquare$


Also see


Sources

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