Let f(x)=∫ 2x (x^2+1)(x^2+3)dx. If f(3)= 1 2( log_e5- log_e6), then f(4) is equal to

Let

$$I=\int \frac{2x}{({x}^{2}+1)({x}^{2}+3)}dx$$I=∫(x2+1)(x2+3)2x​dx

Put ${x}^{2}=t\Rightarrow 2xdx=dt$x2=t⇒2xdx=dt

$$I=\int \frac{1}{(t+1)(t+3)}dt$$I=∫(t+1)(t+3)1​dt

$$\Rightarrow I=\frac{1}{2}\int \frac{2}{(t+1)(t+3)}dt$$⇒I=21​∫(t+1)(t+3)2​dt

$$\Rightarrow I=\frac{1}{2}\int (\frac{1}{t+1}-\frac{1}{t+3})dt$$⇒I=21​∫(t+11​−t+31​)dt

$$\Rightarrow I=\frac{1}{2}[\mathrm{ln}(t+1)-\mathrm{ln}(t+3)]+C$$⇒I=21​[ln(t+1)−ln(t+3)]+C

$$\Rightarrow f(x)=\frac{1}{2}[\mathrm{ln}({x}^{2}+1)-\mathrm{ln}({x}^{2}+3)]+C$$⇒f(x)=21​[ln(x2+1)−ln(x2+3)]+C

Put $x=3$x=3, then

$$\frac{1}{2}[\mathrm{ln}5-\mathrm{ln}6]=\frac{1}{2}[\mathrm{ln}10-\mathrm{ln}12]+C$$21​[ln5−ln6]=21​[ln10−ln12]+C

$$\frac{1}{2}[\mathrm{ln}5-\mathrm{ln}6]=\frac{1}{2}[\mathrm{ln}2+\mathrm{ln}5-\mathrm{ln}2-\mathrm{ln}6]+C$$21​[ln5−ln6]=21​[ln2+ln5−ln2−ln6]+C

$$\Rightarrow C=0$$⇒C=0

So,

$$f(x)=\frac{1}{2}[\mathrm{ln}({x}^{2}+1)-\mathrm{ln}({x}^{2}+3)]$$f(x)=21​[ln(x2+1)−ln(x2+3)]

$\Rightarrow f(4)=\frac{1}{2}(\mathrm{ln}17-\mathrm{ln}19)$⇒f(4)=21​(ln17−ln19) or $f(4)=\frac{1}{2}({\mathrm{log}}_{e}17-{\mathrm{log}}_{e}19)$f(4)=21​(loge​17−loge​19)