Search any question & find its solution
Question:
Answered & Verified by Expert
$\int_0^{\infty}\left(x^{12}+x^{-12}\right) \frac{\log x}{x} d x=$
Options:
Solution:
1606 Upvotes
Verified Answer
The correct answer is:
$0$
$I=\int_0^{\infty}\left(x^{12}+x^{-12}\right) \frac{\log x}{x} d x$
Put $t=\frac{1}{x} \Rightarrow d t=-\frac{1}{x^2} d x$
$\Rightarrow \quad-x d t=\frac{1}{x} d x \Rightarrow-\frac{d t}{t}=\frac{1}{x} d x$
$\therefore \quad I=\int_{\infty}^0\left(t^{-12}+t^{12}\right)(-\log t)\left(\frac{-d t}{t}\right)$
$\begin{aligned} & =\int_{\infty}^0\left(t^{-12}+t^{12}\right)\left(\frac{\log t}{t}\right) d t \\ & =-\int_0^{\infty}\left(t^{-12}+t^{12}\right)\left(\frac{\log t}{t}\right) d t\end{aligned}$
$\begin{aligned} I & =-\int_0^{\infty}\left(x^{-12}+x^{12}\right)\left(\frac{\log x}{x}\right) d x \\ 2 I & =0 \Rightarrow I=0\end{aligned}$
Put $t=\frac{1}{x} \Rightarrow d t=-\frac{1}{x^2} d x$
$\Rightarrow \quad-x d t=\frac{1}{x} d x \Rightarrow-\frac{d t}{t}=\frac{1}{x} d x$
$\therefore \quad I=\int_{\infty}^0\left(t^{-12}+t^{12}\right)(-\log t)\left(\frac{-d t}{t}\right)$
$\begin{aligned} & =\int_{\infty}^0\left(t^{-12}+t^{12}\right)\left(\frac{\log t}{t}\right) d t \\ & =-\int_0^{\infty}\left(t^{-12}+t^{12}\right)\left(\frac{\log t}{t}\right) d t\end{aligned}$
$\begin{aligned} I & =-\int_0^{\infty}\left(x^{-12}+x^{12}\right)\left(\frac{\log x}{x}\right) d x \\ 2 I & =0 \Rightarrow I=0\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.