Practicing Success
Practicing Success
CUET Preparation Today
$∫e^x\left(\frac{1-x}{1+x^2}\right)^2 $ is equal to |
$\frac{e^x}{1+x^2}+C$ $\frac{-e^x}{1+x^2}+C$ $\frac{e^x}{(1+x^2)^2}+C$ $\frac{-e^x}{(1+x^2)^2}+C$ |
$\frac{e^x}{1+x^2}+C$ |
