For \(x \gt -1\), \(\int x\log(1 + x)\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}(x^2 - 1)\log(1 + x) - \frac{1}{4}x^2 + \frac{1}{2}x + C\)
With \(\log(1 + x)\) as the first function and x as the second, integration by parts gives
\( \displaystyle \int x\log(1 + x)\,dx = \frac{x^2}{2}\log(1 + x) - \int \frac{x^2}{2} \cdot \frac{1}{1 + x}\,dx \)
The fraction is improper, so divide \(x^2\) by \(1 + x\):
\( \displaystyle \frac{x^2}{1 + x} = x - 1 + \frac{1}{1 + x} \)
\( \displaystyle \int \frac{x^2}{1 + x}\,dx = \frac{x^2}{2} - x + \log(1 + x) \)
Putting this back:
\( \displaystyle \int x\log(1 + x)\,dx = \frac{x^2}{2}\log(1 + x) - \frac{1}{2}\left(\frac{x^2}{2} - x + \log(1 + x)\right) + C \)
\( \displaystyle = \frac{1}{2}(x^2 - 1)\log(1 + x) - \frac{1}{4}x^2 + \frac{1}{2}x + C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{2}x^2\log(1 + x) - \frac{1}{4}x^2 + \frac{1}{2}x + C\) — Incorrect. The term \(-\frac{1}{2}\log(1 + x)\) from \(\int \frac{dx}{1 + x}\) is missing.
Option (3) \(\displaystyle\frac{1}{2}(x^2 - 1)\log(1 + x) + \frac{1}{4}x^2 - \frac{1}{2}x + C\) — Incorrect. The signs of the last two terms are reversed; the minus sign in front of the second integral was not applied to them.
Option (4) \(\displaystyle\frac{1}{2}(x^2 + 1)\log(1 + x) - \frac{1}{4}x^2 - \frac{1}{2}x + C\) — Incorrect. Its derivative is \(x\log(1 + x) - \frac{x}{1 + x}\), not \(x\log(1 + x)\).