STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

$=16 \int \limits_1^2 \frac{d x}{x^3 x^4\left(1+\frac{2}{x^2}\right)^2}$

Let, $1+\frac{2}{x^2}=t \Rightarrow \frac{-4}{x^3} d x=d t$

$I=-4 \int \limits_3^{\frac{3}{2}} \frac{d t}{\left(\frac{2}{t-1}\right)^2 t^2}$

$I=-4 \int \limits_3^{\frac{3}{2}}\left(\frac{t-1}{2}\right)^2 \frac{d t}{t^2}$

$I=-\frac{4}{4} \int \limits_3^{\frac{3}{2}}\left(1-\frac{2}{t}+\frac{1}{t^2}\right) d t$

$I=-1\left[t-2 \ell n|t|-\frac{1}{t}\right]_3^{\frac{3}{2}}$

$I=-1\left[\left(\frac{3}{2}-2 \ell n \frac{3}{2}-\frac{2}{3}\right)-\left(3-2 \ell n 3-\frac{1}{3}\right)\right]$

$I=-1\left[2 \ell n 2-\frac{11}{6}\right]$

$I=\frac{11}{6}-\ell n 4$