Q113 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
$$
\int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x=
$$
- Correct answerA$\frac{1}{9}\left(1+\frac{1}{x^2}\right)^{\frac{3}{2}}\left[2-3 \log \left(1+\frac{1}{x^2}\right)\right\rfloor+c$
- B$\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{\frac{1}{2}}\left[6-\log \left(1+\frac{1}{x^2}\right)^2\right]+c$
- C$\frac{1}{9}\left(1+\frac{1}{x^2}\right)\left[3-2 \log \left(1+\frac{1}{x^2}\right)^{\frac{1}{2}}\right]+c$
- D$\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{\frac{3}{2}}\left[3+\log \left(1+\frac{1}{x^2}\right)\right]+c$
Solution
$$
\begin{array}{r}
\text { } I=\int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x \\
=\int \frac{1}{x^3} \sqrt{1+\frac{1}{x^2}} \log \left(1+\frac{1}{x^2}\right) d x
\end{array}
$$
Let $1+\frac{1}{x^2}=t^2 \Rightarrow-2 \frac{d x}{x^3}=2 t d t$
So,
$$
I=-\int t^2 \log t^2 d t
$$
$$
=-2 \int t^2 \log t d t=-2\left[\frac{t^3}{3} \log t-\int\left(\frac{t^3}{3} \times \frac{1}{t}\right) d t\right]
$$
$$
=-\frac{2}{3} t^3 \log t+\frac{2}{9} t^3+c=\frac{1}{9} t^3\left[2-3 \log t^2\right]+c
$$
$$
=\frac{1}{9}\left(1+\frac{1}{x^2}\right)^{3 / 2}\left[2-3 \log \left(1+\frac{1}{x^2}\right)\right]+c
$$
Hence, the correct answer is option: 'A'.