Q114 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics

$\int_0^{\infty}\left(x^{12}+x^{-12}\right) \frac{\log x}{x} d x=$

  • Correct answer
    A
    $0$
  • B
    $1$
  • C
    $\log 2$
  • D
    $e^2$

Solution

$I=\int_0^{\infty}\left(x^{12}+x^{-12}\right) \frac{\log x}{x} d x$ Put $t=\frac{1}{x} \Rightarrow d t=-\frac{1}{x^2} d x$ $\Rightarrow \quad-x d t=\frac{1}{x} d x \Rightarrow-\frac{d t}{t}=\frac{1}{x} d x$ $\therefore \quad I=\int_{\infty}^0\left(t^{-12}+t^{12}\right)(-\log t)\left(\frac{-d t}{t}\right)$ $\begin{aligned} & =\int_{\infty}^0\left(t^{-12}+t^{12}\right)\left(\frac{\log t}{t}\right) d t \\ & =-\int_0^{\infty}\left(t^{-12}+t^{12}\right)\left(\frac{\log t}{t}\right) d t\end{aligned}$ $\begin{aligned} I & =-\int_0^{\infty}\left(x^{-12}+x^{12}\right)\left(\frac{\log x}{x}\right) d x \\ 2 I & =0 \Rightarrow I=0\end{aligned}$

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

Relevant Questions