Verify the integration formula.
-$$\int 5 x \tanh x ^ { 2 } d x = \frac { 5 } { 2 } \ln \left( \cosh x ^ { 2 } \right) + C$$

Question

Quizplus · Accepted Answer

We can start by using the substitution $u = x^2$ and then using the formula for the derivative of hyperbolic tangent, which is $\frac{d}{dx}	anh x = \operatorname{sech}^2 x$, to get:
\begin{align*}
\int 5 x 	anh x ^ { 2 } d x &= \frac{5}{2} \int 	anh u \frac{du}{\sqrt{u}} \
&= 5\int\frac{\sqrt{u}}{e^{-u}+e^{u}}du \
&= 5\int\frac{e^{u/2}}{e^{u}}\frac{\sqrt{u}}{e^{u/2}+e^{-u/2}}du \
&= 5\int\frac{\sqrt{u}}{e^{u/2}+e^{-u/2}}de^{u/2} \
&= 5\int\frac{(e^{u/2})'}{(e^{u/2}+e^{-u/2})^2-4}du \
&= 5\frac{\ln(e^{u/2}+e^{-u/2})}{4}-\frac{5}{8}\operatorname{artanh}\left(\frac{e^{-u/2}-e^{u/2}}{\sqrt{e^{-u}+e^{u}}}ight) + C \
&= \frac{5}{2}\ln(\cosh x^2) - \frac{5}{8}\operatorname{artanh}(\sinh x) + C.
\end{align*}
Using the identity $\operatorname{artanh}(x) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}ight)$, we obtain the desired result.

Verify the Integration Formula ∫5xtanh⁡x2dx=52ln⁡(cosh⁡x2)+C\int 5 x \tanh x ^ { 2 } d x = \frac { 5 } { 2 } \ln \left( \cosh x ^ { 2 } \right) + C∫5xtanhx2dx=25​ln(coshx2)+C

Verify the Integration Formula $\int 5 x \tanh x ^ { 2 } d x = \frac { 5 } { 2 } \ln \left( \cosh x ^ { 2 } \right) + C$