Q58 / 150 · MHT CET 2020 (20 Oct Shift 1) · Mathematics

$\int_{1}^{28} \frac{\mathrm{d} x}{x(1+\log x)^{2}}=$

  • A
    $\cdot \log 2$
  • B
    $1+\log 2$
  • Correct answer
    C
    $\frac{\log 2}{(1+\log 2)}$
  • D
    $\frac{1}{(1+\log 2)}$

Solution

$I=\int_{1}^{2} \frac{d x}{x(1+\log x)}$ Put $\quad 1+\log x=t \Rightarrow \frac{1}{x} d x=d t$ When $x=1, t=1$ and when $x=2, t=1+\log 2$ $\begin{aligned} I &=\int_{1}^{1+\log 2} \frac{d t}{t}=[\log t]_{1}^{1+\log 2}=\log (1+\log 2)-\log 1 \\ &=\log (1+\log 2) \end{aligned}$

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

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