\(\int\displaystyle\frac{dx}{(x+1)\left(x^2+1\right)}\) is equal to (C is a constant of integration)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{2}\log|x+1|-\frac{1}{4}\log\left(x^2+1\right)+\frac{1}{2}\tan^{-1}x+C\)
The factor \(x^2+1\) has no real roots, so it takes a linear numerator in the partial fractions:
\( \displaystyle \frac{1}{(x+1)\left(x^2+1\right)} = \frac{A}{x+1}+\frac{Bx+D}{x^2+1} \)
\( 1 = A\left(x^2+1\right)+(Bx+D)(x+1) \)
Putting x = −1:
\( \displaystyle 1 = 2A,\quad A = \frac{1}{2} \)
Comparing the coefficients of \(x^2\) and the constant terms:
\( \displaystyle 0 = A+B,\quad B = -\frac{1}{2} \)
\( \displaystyle 1 = A+D,\quad D = \frac{1}{2} \)
So
\( \displaystyle I = \frac{1}{2}\int\frac{dx}{x+1}-\frac{1}{2}\int\frac{x}{x^2+1}\,dx+\frac{1}{2}\int\frac{dx}{x^2+1} \)
Since \(\int\frac{x}{x^2+1}\,dx = \frac{1}{2}\log\left(x^2+1\right)\):
\( \displaystyle I = \frac{1}{2}\log|x+1|-\frac{1}{4}\log\left(x^2+1\right)+\frac{1}{2}\tan^{-1}x+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{2}\log|x+1|-\frac{1}{2}\log\left(x^2+1\right)+\frac{1}{2}\tan^{-1}x+C\) — Incorrect. It misses the factor \(\frac{1}{2}\) in \(\int\frac{x}{x^2+1}\,dx = \frac{1}{2}\log\left(x^2+1\right)\).
Option (2) \(\displaystyle\frac{1}{2}\log|x+1|+\frac{1}{4}\log\left(x^2+1\right)-\frac{1}{2}\tan^{-1}x+C\) — Incorrect. The signs of B and D are reversed; B is \(-\frac{1}{2}\) and D is \(+\frac{1}{2}\).
Option (3) \(\displaystyle\frac{1}{2}\log|x+1|-\frac{1}{4}\log\left(x^2+1\right)-\frac{1}{2}\tan^{-1}x+C\) — Incorrect. D is \(+\frac{1}{2}\), so the \(\tan^{-1}x\) term is positive.