Question

$\int \frac{2x\log (1+x^2)}{1+x^2}dx$

• A. $\log (1+x^2)+c$
• B. $\frac{\left[\log \right(1+x^2){]}^2}{2}+c$
• C. $2\log (1+x^2)+c$
• D. none of these

Answer 1

The correct option is B $\frac{\left[\log \right(1+x^2){]}^2}{2}+c$

Now,

$\int \frac{2x\log (1+x^2)}{1+x^2}dx$

Let $\log (1+x^2)=z$ and $\frac{2x}{1+x^2}dx=dz$.

Using these in the above integration we get,

$=\int zdz$

$=\frac{z^2}{2}+c$ [ Where $c$ is integrating constant]

$=\frac{\left[\log \right(1+x^2){]}^2}{2}+c$