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 this result by differentiating the given formula and checking if it matches the integrand. Using the derivative of inverse hyperbolic tangent, we have:

\begin{align*}
\frac{d}{dx}\left[x	anh^{-1}x + \frac{1}{2}\ln(1-x^2)ight] &= \frac{d}{dx}\left[x	anh^{-1}xight] + \frac{d}{dx}\left[\frac{1}{2}\ln(1-x^2)ight]\
&= 	anh^{-1}x + x\frac{1}{1-x^2}\cdot (-2x)\
&= 	anh^{-1}x + \frac{-2x^2}{1-x^2}\
&= 	anh^{-1}x - \frac{2x^2-2+2}{1-x^2}\
&= 	anh^{-1}x - \frac{2}{1-x^2} + 2\cdot\frac{1}{1-x^2}\
&= 	anh^{-1}x - \frac{2}{1-x^2} + \frac{2}{1-x^2}\
&= 	anh^{-1}x
\end{align*}

Therefore, the derivative of the given formula matches the integrand, and so the formula is verified.

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