If $\int\limits_0^1 \frac{e^x}{1+x} d x=K$, then $\int\limits_0^1 \frac{e^x}{(1+x)^2} d x$ is equal to :
Answer & explanation
Correct answer: option 2
so $I=\int\limits_0^1 \frac{e^x}{1+x} d x$
we get $I = \left[\frac{e^x}{1+x}\right]_0^1+\int\limits_0 \frac{e^x}{1+x^2} d x=K$ (as given)
So $\int\limits_0^1 \frac{e^x}{(1+x)^2} d x=K-\left[\frac{e x}{1+x}\right]_0^1$
$\Rightarrow \int\limits_0^1 \frac{e^x d x}{(1+x)^2}=K-\left[\frac{e}{2}-1\right]$
$\Rightarrow \int\limits_0^1 \frac{e^x}{(1+x)^2} d x=K-\frac{e}{2}+1$