\(\int\displaystyle\frac{x^2\tan^{-1}\left(x^3\right)}{1+x^6}\,dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{6}\left(\tan^{-1}x^3\right)^2+C\)
The rest of the integrand is, up to a constant, the derivative of \(\tan^{-1}\left(x^3\right)\), so substitute
\( \displaystyle t = \tan^{-1}\left(x^3\right),\quad dt = \frac{3x^2}{1+x^6}\,dx \)
Then
\( \displaystyle \int\frac{x^2\tan^{-1}\left(x^3\right)}{1+x^6}\,dx = \frac{1}{3}\int t\,dt \)
\( \displaystyle = \frac{1}{3}\times\frac{t^2}{2}+C = \frac{1}{6}\left(\tan^{-1}x^3\right)^2+C \)
Why the other options are incorrect
Option (1) \(\frac{1}{3}\left(\tan^{-1}x^3\right)^2+C\) — Incorrect. It forgets the \(\frac{1}{2}\) from \(\int t\,dt = \frac{t^2}{2}\).
Option (3) \(\frac{1}{2}\left(\tan^{-1}x^3\right)^2+C\) — Incorrect. It forgets the factor \(\frac{1}{3}\) from \(dt = \frac{3x^2}{1+x^6}\,dx\).
Option (4) \(\frac{1}{3}\tan^{-1}\left(x^3\right)+C\) — Incorrect. That is the integral of \(\frac{x^2}{1+x^6}\) alone; the extra factor \(\tan^{-1}\left(x^3\right)\) is left out.