\(\int\displaystyle\frac{x}{(x+1)^2}\,dx\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\log|x+1|+\frac{1}{x+1}+C\)
Writing the numerator in terms of the repeated factor \(x+1\):
\( \displaystyle \frac{x}{(x+1)^2} = \frac{(x+1)-1}{(x+1)^2} = \frac{1}{x+1}-\frac{1}{(x+1)^2} \)
Since \(\int\frac{dx}{(x+1)^2} = -\frac{1}{x+1}\):
\( \displaystyle \int\frac{x}{(x+1)^2}\,dx = \log|x+1|+\frac{1}{x+1}+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\log|x+1|-\frac{1}{x+1}+C\) — Incorrect. \(\int\frac{dx}{(x+1)^2} = -\frac{1}{x+1}\), so subtracting it gives \(+\frac{1}{x+1}\).
Option (2) \(\displaystyle -\frac{1}{x+1}+C\) — Incorrect. Its derivative is \(\frac{1}{(x+1)^2}\), which has numerator 1, not x.
Option (4) \(\displaystyle\frac{1}{2}\log\left|x^2+2x+1\right|+C\) — Incorrect. This equals \(\log|x+1|\), whose derivative is \(\frac{1}{x+1}\).