Question

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

• A. $\frac{1}{3}{(1+\frac{1}{x^2})}^{\frac{3}{2}}[\log (1+\frac{1}{x^2})+\frac{2}{3}]+C$
• B. $-\frac{1}{3}{(1+\frac{1}{x^2})}^{\frac{3}{2}}[\log (1+\frac{1}{x^2})-\frac{2}{3}]+C$
• C. $\frac{2}{3}{(1+\frac{1}{x^2})}^{\frac{1}{2}}[\log (1+\frac{1}{x^2})+\frac{2}{3}]+C$
• D. None of these

Answer 1

The correct option is B $-\frac{1}{3}{(1+\frac{1}{x^2})}^{\frac{3}{2}}[\log (1+\frac{1}{x^2})-\frac{2}{3}]+C$

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

using $\{\log x^n=n\log x\log \left(\frac{x}{y}\right)=\log x-\log y\}$

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

$\int \sqrt{1+\frac{1}{x^2}}\frac{\left[\log (1+\frac{1}{x^2})\right]}{x^3}dx$

let $1+\frac{1}{x^2}=t$

$-\frac{2}{x^3}dx=dt\Rightarrow \frac{dx}{x^3}=-\frac{dt}{2}$

Now we can write integer as-

$=-\frac{1}{2}\int \sqrt{t}(\log .t)dt$ using integration by parts

$=-\frac{1}{2}\left[logt\int \sqrt{t}dt-\int \frac{d}{dt}\left(\log t\right)\int \sqrt{t}dt\int dt\right]$

$=-\frac{1}{2}\left[\right(\log t\left)\frac{2}{3}\right(t{)}^{3/2}-\frac{2}{3}\int \sqrt{t}dt]$

$=-\frac{1}{2}\left[\frac{2}{3}\right(t{)}^{3/2}\log \left(t\right)-\frac{4}{9}\left(t{)}^{3/2}\right]+C$

$=-\frac{1}{3}(t{)}^{3/2}[\log t+-\frac{2}{3}]+C$ [replaced

$t=1+\frac{1}{x^2}$]

$=-\frac{1}{3}{(1+\frac{1}{x^2})}^{\frac{3}{2}}[\log (1+\frac{1}{x^2})-\frac{2}{3}]+C$ Answer