Q99 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics

$\int \dfrac{\log x}{(1 + x)^2}\, dx$

  • A
    $\dfrac{\log x}{x + 1} - \log\left|\dfrac{x}{x + 1}\right| + C$
  • B
    $-\dfrac{\log x}{x + 1} - \log\left|\dfrac{x}{x + 1}\right| + C$
  • Correct answer
    C
    $-\dfrac{\log x}{x + 1} + \log\left|\dfrac{x}{x + 1}\right| + C$
  • D
    $\dfrac{\log x}{x + 1} + \log\left|\dfrac{x + 1}{x}\right| + C$

Solution

Using integration by parts:

$\int \dfrac{\log x}{(1 + x)^2} dx$

Let $u = \log x$ and $dv = \dfrac{1}{(1 + x)^2} dx$

$du = \dfrac{1}{x} dx$ and $v = -\dfrac{1}{1 + x}$

$I = -\dfrac{\log x}{1 + x} - \int \left(-\dfrac{1}{1 + x}\right) \dfrac{1}{x} dx$

$I = -\dfrac{\log x}{1 + x} + \int \dfrac{1}{x(1 + x)} dx$

Using partial fractions:

$\dfrac{1}{x(1 + x)} = \dfrac{1}{x} - \dfrac{1}{1 + x}$

$I = -\dfrac{\log x}{1 + x} + \int \left(\dfrac{1}{x} - \dfrac{1}{1 + x}\right) dx$

$I = -\dfrac{\log x}{1 + x} + \log|x| - \log|1 + x| + C$

$I = -\dfrac{\log x}{x + 1} + \log\left|\dfrac{x}{x + 1}\right| + C$

Answer: $-\dfrac{\log x}{x + 1} + \log\left|\dfrac{x}{x + 1}\right| + C$

Hence, the correct answer is option: 'C'.

Relevant Questions