Consider the following statements, where C is a constant of integration:
(A) \(\int\displaystyle\frac{dx}{x\left(x^2+1\right)} = \log|x|-\frac{1}{2}\log\left(x^2+1\right)+C\)
(B) \(\int\displaystyle\frac{x-1}{x(x+1)}\,dx = 2\log|x+1|-\log|x|+C\)
(C) \(\int\displaystyle\frac{dx}{x^2(x+1)} = \log\left|\frac{x+1}{x}\right|+\frac{1}{x}+C\)
(D) \(\int\displaystyle\frac{x^3}{x^2+1}\,dx = \frac{x^2}{2}-\log\left(x^2+1\right)+C\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
Each integrand is split into partial fractions (after division where the fraction is improper):
\( \displaystyle \frac{1}{x\left(x^2+1\right)} = \frac{1}{x}-\frac{x}{x^2+1} \)
\( \displaystyle \frac{x-1}{x(x+1)} = -\frac{1}{x}+\frac{2}{x+1} \)
\( \displaystyle \frac{1}{x^2(x+1)} = -\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1} \)
\( \displaystyle \frac{x^3}{x^2+1} = x-\frac{x}{x^2+1} \)
Integrating, with \(\int\frac{x}{x^2+1}\,dx = \frac{1}{2}\log\left(x^2+1\right)\) and \(\int\frac{dx}{x^2} = -\frac{1}{x}\):
So the four integrals are \(\log|x|-\frac{1}{2}\log\left(x^2+1\right)\), \(2\log|x+1|-\log|x|\), \(\log\left|\frac{x+1}{x}\right|-\frac{1}{x}\) and \(\frac{x^2}{2}-\frac{1}{2}\log\left(x^2+1\right)\), each \(+C\).
Checking the statements
(A) \(\int\frac{dx}{x\left(x^2+1\right)} = \log|x|-\frac{1}{2}\log\left(x^2+1\right)+C\) — Correct. It integrates \(\frac{1}{x}-\frac{x}{x^2+1}\).
(B) \(\int\frac{x-1}{x(x+1)}\,dx = 2\log|x+1|-\log|x|+C\) — Correct. It integrates \(\frac{2}{x+1}-\frac{1}{x}\).
(C) \(\int\frac{dx}{x^2(x+1)} = \log\left|\frac{x+1}{x}\right|+\frac{1}{x}+C\) — Incorrect. \(\int\frac{dx}{x^2} = -\frac{1}{x}\), so the last term is \(-\frac{1}{x}\).
(D) \(\int\frac{x^3}{x^2+1}\,dx = \frac{x^2}{2}-\log\left(x^2+1\right)+C\) — Incorrect. The log term is \(\frac{1}{2}\log\left(x^2+1\right)\).