Consider the following statements, where C is a constant of integration and \(x \gt 0\) in (B):
(A) \(\int xe^{-x}\,dx = -(x+1)e^{-x}+C\)
(B) \(\int x^3\log x\,dx = \displaystyle\frac{x^4}{4}\log x-\frac{x^4}{16}+C\)
(C) \(\int x^2e^x\,dx = e^x\left(x^2-2x-2\right)+C\)
(D) \(\int x\sin\displaystyle\frac{x}{2}\,dx = -2x\cos\frac{x}{2}+4\sin\frac{x}{2}+C\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
Each integral is found by integration by parts, \(\int u\,v\,dx = u\int v\,dx-\int\left(u'\int v\,dx\right)dx\), with the polynomial or log as u.
\( \int xe^{-x}\,dx = -xe^{-x}-e^{-x} = -(x+1)e^{-x} \)
\( \displaystyle \int x^3\log x\,dx = \frac{x^4}{4}\log x-\int\frac{x^3}{4}\,dx = \frac{x^4}{4}\log x-\frac{x^4}{16} \)
\( \int x^2e^x\,dx = x^2e^x-2\left(xe^x-e^x\right) = e^x\left(x^2-2x+2\right) \)
\( \displaystyle \int x\sin\frac{x}{2}\,dx = -2x\cos\frac{x}{2}+2\int\cos\frac{x}{2}\,dx = -2x\cos\frac{x}{2}+4\sin\frac{x}{2} \)
So (A), (B) and (D) match these results, while \(\int x^2e^x\,dx = e^x\left(x^2-2x+2\right)+C\).
Checking the statements
(A) \(\int xe^{-x}\,dx = -(x+1)e^{-x}+C\) — Correct. Its derivative is \(-e^{-x}+(x+1)e^{-x} = xe^{-x}\).
(B) \(\int x^3\log x\,dx = \frac{x^4}{4}\log x-\frac{x^4}{16}+C\) — Correct. Its derivative is \(x^3\log x+\frac{x^3}{4}-\frac{x^3}{4} = x^3\log x\).
(C) \(\int x^2e^x\,dx = e^x\left(x^2-2x-2\right)+C\) — Incorrect. The last term is \(+2\); the given result has derivative \(e^x\left(x^2-4\right)\).
(D) \(\int x\sin\frac{x}{2}\,dx = -2x\cos\frac{x}{2}+4\sin\frac{x}{2}+C\) — Correct. Its derivative is \(-2\cos\frac{x}{2}+x\sin\frac{x}{2}+2\cos\frac{x}{2} = x\sin\frac{x}{2}\).