Question

Evaluate \( \displaystyle \int_{0}^{1} \tan^{-1}\!\left(\frac{2x}{1-x^2}\right)\,dx \)

• A. \( \pi - \log 2 \)
• B. \( \dfrac{\pi}{2} - \log 2 \)
• C. \( \pi + \log 2 \)
• D. \( \dfrac{\pi}{2} + \log 2 \)

Hint:

Whenever you see \( \tan^{-1}\!\left(\frac{2x}{1-x^2}\right) \), try using the double-angle identity of inverse tangent.

Answer 1

The Correct Option is B

Step 1: Use the identity for inverse tangent.
We know that \[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x), \quad |x|<1 \] Step 2: Rewrite the integral. 
\[ \int_{0}^{1} \tan^{-1}\left(\frac{2x}{1-x^2}\right)\,dx = 2\int_{0}^{1} \tan^{-1}(x)\,dx \] Step 3: Evaluate \( \int \tan^{-1}(x)\,dx \). 
\[ \int \tan^{-1}(x)\,dx = x\tan^{-1}(x) - \frac{1}{2}\ln(1+x^2) \] Step 4: Apply the limits. 
\[ 2\left[ x\tan^{-1}(x) - \frac{1}{2}\ln(1+x^2) \right]_{0}^{1} \] \[ = 2\left[ \frac{\pi}{4} - \frac{1}{2}\ln 2 \right] \] Step 5: Simplify. 
\[ = \frac{\pi}{2} - \ln 2 \] Step 6: Conclusion. 
The value of the given integral is \[ \frac{\pi}{2} - \log 2 \]