Let f (x) = ∫ 2x ( x ^ 2 + 1)( x ^ 2 + 3) dx . If f(3) = (1)/(2)( _...

Let

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

Put ${x}^{2} = t ⇒ 2xdx = dt$

$I = ∫ \dfrac{1}{\left(t + 1\right)\left(t + 3\right)}dt$

$⇒ I = \dfrac{1}{2} ∫ \dfrac{2}{\left(t + 1\right)\left(t + 3\right)}dt$

$⇒ I = \dfrac{1}{2} ∫ \left(\dfrac{1}{t + 1} - \dfrac{1}{t + 3}\right)dt$

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

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

Put $x = 3$, then

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

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

$⇒ C = 0$

So,

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

$⇒ f\left(4\right) = \dfrac{1}{2}\left(\ln 17 - \ln 19\right)$ or $f\left(4\right) = \dfrac{1}{2}\left(\log_{e}17 - \log_{e}19\right)$