Match List-I with List-II (\(x \gt 0\), C is a constant of integration)
| List-I | List-II | ||
| (A) | \(\int\tan^{-1}(2x)\,dx\) | (I) | \(\left(x^2+1\right)\tan^{-1}x-x+C\) |
| (B) | \(\int 2x\tan^{-1}x\,dx\) | (II) | \(-x^2\cos x+2x\sin x+2\cos x+C\) |
| (C) | \(\int x^2\sin x\,dx\) | (III) | \(\displaystyle\frac{x^2}{2}(\log x)^2-\frac{x^2}{2}\log x+\frac{x^2}{4}+C\) |
| (D) | \(\int x(\log x)^2\,dx\) | (IV) | \(\displaystyle x\tan^{-1}(2x)-\frac{1}{4}\log\left(1+4x^2\right)+C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
All four are done by integration by parts, taking the inverse trigonometric, polynomial or log factor as the first function by the usual order of preference.
(A) \(\int\tan^{-1}(2x)\,dx\)
This matches (IV) \(x\tan^{-1}(2x)-\frac{1}{4}\log\left(1+4x^2\right)+C\).
\( \displaystyle \int\tan^{-1}(2x)\,dx = x\tan^{-1}(2x)-\int\frac{2x}{1+4x^2}\,dx \)
\( \displaystyle = x\tan^{-1}(2x)-\frac{1}{4}\log\left(1+4x^2\right)+C \)
(B) \(\int 2x\tan^{-1}x\,dx\)
This matches (I) \(\left(x^2+1\right)\tan^{-1}x-x+C\).
\( \displaystyle \int 2x\tan^{-1}x\,dx = x^2\tan^{-1}x-\int\frac{x^2}{1+x^2}\,dx \)
\( \displaystyle = x^2\tan^{-1}x-\left(x-\tan^{-1}x\right)+C = \left(x^2+1\right)\tan^{-1}x-x+C \)
(C) \(\int x^2\sin x\,dx\)
This matches (II) \(-x^2\cos x+2x\sin x+2\cos x+C\).
\( \displaystyle \int x^2\sin x\,dx = -x^2\cos x+\int 2x\cos x\,dx \)
\( = -x^2\cos x+2x\sin x+2\cos x+C \)
(D) \(\int x(\log x)^2\,dx\)
This matches (III) \(\frac{x^2}{2}(\log x)^2-\frac{x^2}{2}\log x+\frac{x^2}{4}+C\).
\( \displaystyle \int x(\log x)^2\,dx = \frac{x^2}{2}(\log x)^2-\int x\log x\,dx \)
\( \displaystyle = \frac{x^2}{2}(\log x)^2-\left(\frac{x^2}{2}\log x-\frac{x^2}{4}\right)+C \)