Question:
If $f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0)$
and $f(0)=0$, then the value of $f(1)$ is:
Correct Option: , 4
Solution:
$f(x)=$
$=\int \frac{5 x^{8}+7 x^{6}}{x^{14}\left(x^{-5}+x^{-7}+2\right)^{2}} d x$
$=\int \frac{5 x^{-6}+7 x^{-8}}{\left(2+x^{-7}+x^{-5}\right)^{2}} d x$
Let $2+x^{-7}+x^{-5}=t$
$\Rightarrow \quad\left(-7 x^{-8}-5 x^{-6}\right) d x=d t$
$\Rightarrow f(x)=\int \frac{-d t}{t^{2}}=\int-t^{-2} d t=t^{-1}+c$
$\Rightarrow f(x)=\frac{1}{2+x^{-7}+x^{-5}}+c, f(0)=0 \Rightarrow c=0$
$\therefore \quad f(1)=\frac{1}{4}$
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