Question

$\int \left[\frac{\sqrt{x^2+1}\left[ln\right(x^2+1)-2lnx]}{x^4}\right]$

• A. $\frac{(x^2-1)\sqrt{x^2-1}}{9x^3}[2-3ln(1+\frac{1}{x^2})]$
• B. $\frac{(x^2-1)\sqrt{x^2-1}}{9x^3}[2+3ln(1+\frac{1}{x^2})]$
• C. $\frac{(x^2+1)\sqrt{x^2+1}}{9x^3}[2-3ln(1+\frac{1}{x^2})]$
• D. $\frac{(x^2-1)\sqrt{x^2+1}}{9x^3}[2+3ln(1+\frac{1}{x^2})]$

Answer 1

Answer: (C)

$\frac{(x^2+1)\sqrt{x^2+1}}{9x^3}[2-3ln(1+\frac{1}{x^2})]$

Let $I=\int \frac{\sqrt{1+x^2}\{\log (1+x^2)-2\log x\}}{x^4}dx$
$=\int \log \left\{\frac{1+x^2}{x^2}\right\}\cdot \frac{\sqrt{1+x^2}}{x^1.x^3}dx$
Substitute $\frac{1+x^2}{x^2}=1+\frac{1}{x^2}=t\Rightarrow -\frac{2}{x^3}dx=dt$
$I=-\frac{1}{2}\int \sqrt{t}.\log tdt=-\frac{1}{2}[\frac{2}{3}t\sqrt{t}\log t-\frac{4}{9}t\sqrt{t}]$
$=\frac{(x^2+1)\sqrt{x^2+1}}{9x^3}[2-3ln(1+\frac{1}{x^2})]$