Let $f(x)=\int \frac{x}{\left(x^2+1\right)\left(x^2+3\right)} d x$. If $f(3)=\frac{1}{4} \log \left(\frac{5}{6}\right)$ then $f(0)=$

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$\because f(x)=\int \frac{x}{\left(x^2+1ight)\left(x^2+3ight)} d x$ Let $x^2=t \Rightarrow 2 x d x=d t$ $\begin{aligned} & 	ext { So, } f(x)=\frac{1}{2} \int \frac{d t}{(t+1)(t+3)} \ & =\frac{1}{4} \int\left(\frac{1}{t+1}-\frac{1}{t+3}ight) d t=\frac{1}{4} \log \left(\frac{1+x^2}{3+x^2}ight)+c \end{aligned}$ $\begin{aligned} & 	ext { Since, } f(3)=\frac{1}{4} \log \left(\frac{5}{6}ight) \Rightarrow \frac{1}{4} \log \left(\frac{5}{6}ight) \ & =\frac{1}{4} \log \left(\frac{10}{12}ight)+c \Rightarrow c=0 \ & \Rightarrow f(x)=\frac{1}{4} \log \left(\frac{1+x^2}{3+x^2}ight) 	ext { So, } f(0)=\frac{1}{4} \log \left(\frac{1}{3}ight)\end{aligned}$

Let $f(x)=\int \frac{x}{\left(x^2+1\right)\left(x^2+3\right)} d x$. If $f(3)=\frac{1}{4} \log…

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