Question

The integral $16 \int\limits_1^2 \frac{d x}{x^3\left(x^2+2\right)^2}$ is equal to

• A. $\frac{11}{6}-\log _e 4$
• B. $\frac{11}{12}+\log _{ e } 4$
• C. $\frac{11}{6}+\log _{ e } 4$
• D. $\frac{11}{12}-\log _e 4$

Answer 1

Answer: (C)

$ I=16 \int\limits_1^2 \frac{d x}{x^3\left(x^2+2\right)^2}$
$ =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]$
$I=\frac{11}{6}-\ell n 4$