Question

$\int \frac{\sqrt{x^2+1}\left[\log \right(x^2+1)-2\log x]}{x^4}dx$ is equal to -

• A. $\frac{1}{3}{(1+\frac{1}{x^2})}^{1/2}[\log (1+\frac{1}{x^2})+\frac{2}{3}]+c$
• B. $-\frac{1}{3}{(1+\frac{1}{x^2})}^{3/2}[\log (1+\frac{1}{x^2})-\frac{2}{3}]+c$
• C. $\frac{2}{3}{(1+\frac{1}{x^2})}^{1/2}[\log (1+\frac{1}{x^2})+\frac{2}{3}]+c$
• D. $Noneofthese$

Answer 1

Answer: (B)

$-\frac{1}{3}{(1+\frac{1}{x^2})}^{3/2}[\log (1+\frac{1}{x^2})-\frac{2}{3}]+c$

$I=\int \frac{\sqrt{x^2+1}(\log (x^2+1)-2\log x)}{x^4}dx$
$=\int \frac{\sqrt{x^2+1}(\log (x^2+1)-\log x^2)}{x^4}dx$
$=\int \frac{\sqrt{x^2+1}\left(\log \left(\frac{x^2+1}{x^2}\right)\right)}{x^4}dx$
$=\int \frac{\sqrt{x^2+1}\left(\log (1+\frac{1}{x^2})\right)}{x^4}dx$
$=\int \frac{\sqrt{\frac{x^2+1}{x^2}}\left(\log (1+\frac{1}{x^2})\right)}{x^3}dx$
$=\int \frac{\sqrt{1+\frac{1}{x^2}}\left(\log (1+\frac{1}{x^2})\right)}{x^3}dx$
$let1+\frac{1}{x^2}=t^2$
$differentiatingbothsides$
$-\frac{2}{x^3}dx=2tdt$
$I=-\int t\log t^2\left(tdt\right)$
$=-2\int t^2\log tdt$
$\mathrm{int}egrationbyparts$
$=-2[\log t\left(\frac{t^3}{3}\right)-\int \frac{t^2}{3}dt][\int uvdx=u\underset{}{\overset{}{\int }}vdx-\iint v\frac{d\left(u\right)}{dx}]$
$=-2[\frac{t^3}{3}\log t-\frac{t^3}{9}]$
$=-2[\frac{1}{3}\log \sqrt{1+\frac{1}{x^3}}{(1+\frac{1}{x^2})}^{3/2}-\frac{1}{9}{(1+\frac{1}{x^2})}^{3/2}]$
$=\frac{-1}{3}{(1+\frac{1}{x^2})}^{3/2}[\log (1+\frac{1}{x^2})-\frac{2}{3}]+c$