\(\int\log\left(1+x^2\right)dx\) is equal to (C is a constant of integration)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(x\log\left(1+x^2\right)-2x+2\tan^{-1}x+C\)
Writing the integrand as \(\log\left(1+x^2\right)\times 1\), we use integration by parts with 1 as the second function:
\( \displaystyle \int\log\left(1+x^2\right)dx = x\log\left(1+x^2\right)-\int x\cdot\frac{2x}{1+x^2}\,dx \)
The remaining fraction is improper, so we divide:
\( \displaystyle \frac{2x^2}{1+x^2} = 2-\frac{2}{1+x^2} \)
\( \displaystyle \int\frac{2x^2}{1+x^2}\,dx = 2x-2\tan^{-1}x \)
Hence
\( \int\log\left(1+x^2\right)dx = x\log\left(1+x^2\right)-2x+2\tan^{-1}x+C \)
Why the other options are incorrect
Option (1) \(x\log\left(1+x^2\right)-2x-2\tan^{-1}x+C\) — Incorrect. Subtracting \(2x-2\tan^{-1}x\) makes the \(\tan^{-1}x\) term positive.
Option (2) \(x\log\left(1+x^2\right)+2x-2\tan^{-1}x+C\) — Incorrect. It adds the second integral instead of subtracting it.
Option (4) \(x\log\left(1+x^2\right)-x+\tan^{-1}x+C\) — Incorrect. It drops the factor 2 from the derivative \(\frac{2x}{1+x^2}\) of \(\log\left(1+x^2\right)\).