Let f x = ∫ 2 x x 2 + 1 x 2 + 3 d x . If f 3 = 1 2 log e 5 - log e 6 , then f 4 is equal to

Let

$I=\int \frac{2x}{\left(x^2+1\right)\left(x^2+3\right)}dx$

Put $x^2=t\Rightarrow 2xdx=dt$

$I=\int \frac{1}{\left(t+1\right)\left(t+3\right)}dt$

$\Rightarrow I=\frac{1}{2}\int \frac{2}{\left(t+1\right)\left(t+3\right)}dt$

$\Rightarrow I=\frac{1}{2}\int \left(\frac{1}{t+1}-\frac{1}{t+3}\right)dt$

$\Rightarrow I=\frac{1}{2}\left[\ln \left(t+1\right)-\ln \left(t+3\right)\right]+C$

$\Rightarrow f\left(x\right)=\frac{1}{2}\left[\ln \left(x^2+1\right)-\ln \left(x^2+3\right)\right]+C$

Put $x=3$, then

$\frac{1}{2}\left[\ln 5-\ln 6\right]=\frac{1}{2}\left[\ln 10-\ln 12\right]+C$

$\frac{1}{2}\left[\ln 5-\ln 6\right]=\frac{1}{2}\left[\ln 2+\ln 5-\ln 2-\ln 6\right]+C$

$\Rightarrow C=0$

So,

$f\left(x\right)=\frac{1}{2}\left[\ln \left(x^2+1\right)-\ln \left(x^2+3\right)\right]$

$\Rightarrow f\left(4\right)=\frac{1}{2}\left(\ln 17-\ln 19\right)$ or $f\left(4\right)=\frac{1}{2}\left({\log }_{e}17-{\log }_{e}19\right)$