Primitive of x by Square of Hyperbolic Tangent of a x

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Theorem

$\ds \int x \tanh^2 a x \rd x = \frac {x^2} 2 - \frac {x \tanh a x} a + \frac 1 {a^2} \ln \size {\cosh a x} + C$


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:

\(\ds u\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds 1\) Derivative of Identity Function


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \tanh^2 a x\)
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds x - \frac {\tanh a x} a\) Primitive of $\tanh^2 a x$


Then:

\(\ds \int x \tanh^2 a x \rd x\) \(=\) \(\ds x \paren {x - \frac {\tanh a x} a} - \int \paren {x - \frac {\tanh a x} a} \times 1 \rd x + C\) Integration by Parts
\(\ds \) \(=\) \(\ds x^2 - \frac {x \tanh a x} a + \int x \rd x + \frac 1 a \int \tanh a x \rd x + C\) Linear Combination of Integrals
\(\ds \) \(=\) \(\ds x^2 - \frac {x \tanh a x} a + \frac {x^2} 2 + \frac 1 a \int \tanh a x \rd x + C\) Primitive of Power
\(\ds \) \(=\) \(\ds x^2 - \frac {x \tanh a x} a + \frac {x^2} 2 + \frac 1 a \frac {\ln \size {\cosh a x} } a + C\) Primitive of $\tanh a x$
\(\ds \) \(=\) \(\ds \frac {x^2} 2 - \frac {x \tanh a x} a + \frac 1 {a^2} \ln \size {\cosh a x} + C\) simplifying

$\blacksquare$

Also see


Sources