Question

$\int\frac{1}{(1+x^{2})\tan^{-1}\left(\frac{1+x}{1-x}\right)}dx$ is equal to

• A. $\log\left|\tan^{-1}\left(\frac{1-x}{1+x}\right)\right|+C$
• B. $\log\left|\tan^{-1}\left(\frac{1+x}{1-x}\right)\right|+C$
• C. $\frac{1}{2}\log\left|\tan^{-1}\left(\frac{1+x}{1-x}\right)\right|+C$
• D. $\frac{1}{4}\log\left|\tan^{-1}\left(\frac{1+x}{1-x}\right)\right|+C$
• E. $\frac{1}{4}\log\left|\tan^{-1}\left(\frac{1+x^{2}}{1-x}\right)\right|+C$

Hint:

Math Tip: When integrating an expression of the form $\int \frac{f'(x)}{f(x)} dx$, the answer is always $\log|f(x)| + C$. If you recognize that $\frac{d}{dx}\left[\tan^{-1}\left(\frac{1+x}{1-x}\right)\right] = \frac{1}{1+x^2}$, you can write the answer immediately without full substitution!

Answer 1

The Correct Option is B

Concept:
Calculus - Integration by Substitution and Inverse Trigonometry.
Step 1: Simplify the inverse trigonometric term.
We know the standard trigonometric identity $\tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta}$. Let $x = \tan\theta$, meaning $\theta = \tan^{-1}x$. Substitute this into the expression: $$ \frac{1+x}{1-x} = \frac{1 + \tan\theta}{1 - \tan\theta} = \tan\left(\frac{\pi}{4} + \theta\right) $$
Step 2: Evaluate the inverse tangent.
Taking the inverse tangent of both sides: $$ \tan^{-1}\left(\frac{1+x}{1-x}\right) = \tan^{-1}\left[\tan\left(\frac{\pi}{4} + \theta\right)\right] $$ $$ \tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \theta $$ Substitute $\theta = \tan^{-1}x$ back into the equation: $$ \tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \tan^{-1}x $$
Step 3: Apply the substitution method ($u$-substitution).
Let $u = \tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \tan^{-1}x$. Differentiate $u$ with respect to $x$: $$ \frac{du}{dx} = 0 + \frac{1}{1+x^2} $$ $$ du = \frac{1}{1+x^2} dx $$
Step 4: Transform and evaluate the integral.
Substitute $u$ and $du$ into the original integral: $$ \int \frac{1}{\tan^{-1}\left(\frac{1+x}{1-x}\right)} \cdot \left(\frac{1}{1+x^2} dx\right) $$ $$ = \int \frac{1}{u} du $$ The integral of $\frac{1}{u}$ is $\log|u| + C$: $$ = \log|u| + C $$
Step 5: Substitute $u$ back in terms of $x$.
$$ = \log\left|\tan^{-1}\left(\frac{1+x}{1-x}\right)\right| + C $$