Verify the integration formula.
-$$\int \tanh ^ { - 1 } x d x = x \tanh ^ { - 1 } x + \frac { 1 } { 2 } \ln \left( 1 - x ^ { 2 } \right) + C$$

Question

Quizplus · Accepted Answer

We can verify the given integration formula by using the substitution rule and the derivative of $	anh^{-1} x$. Let $u = 	anh^{-1} x$, then $x = 	anh u$, and $dx = 	ext{sech}^2 u du$. Therefore,

\begin{align*}
\int 	anh^{-1} x dx &= \int u 	ext{sech}^2 u du \
&= u	anh u - \int 	anh u d(	anh u) \
&= u 	anh u - \int \frac{d(\cosh u)}{\cosh u} \
&= u	anh u - \ln|\cosh u| + C \
&= x	anh^{-1} x - \ln \left|\cosh(	anh^{-1} x)ight| + C \
&= x	anh^{-1} x - \ln \left|\frac{1}{\sqrt{1-x^2}}ight| + C \
&= x	anh^{-1} x + \frac{1}{2}\ln(1-x^2) + C
\end{align*}

Therefore, the integration formula is verified and the answer is A.

Verify the Integration Formula ∫tanh⁡−1xdx=xtanh⁡−1x+12ln⁡(1−x2)+C\int \tanh ^ { - 1 } x d x = x \tanh ^ { - 1 } x + \frac { 1 } { 2 } \ln \left( 1 - x ^ { 2 } \right) + C∫tanh−1xdx=xtanh−1x+21​ln(1−x2)+C

Verify the Integration Formula $\int \tanh ^ { - 1 } x d x = x \tanh ^ { - 1 } x + \frac { 1 } { 2 } \ln \left( 1 - x ^ { 2 } \right) + C$