Question

If $f\left(x\right)=\int \frac{2x^5+5x^4}{{(4+2x+3x^5)}^2}dx,(x\ge 0)$ and $f\left(0\right)=0,$ then the value of $72\cdot f\left(1\right)$ is

Answer 1

Given : $f\left(x\right)=\int \frac{2x^5+5x^4}{{(4+2x+3x^5)}^2}dx$
Taking $x^{10}$ common from numerator and denominator, we get
$f\left(x\right)=\int \frac{2x^{-5}+5x^{-6}}{{(4x^{-5}+2x^{-4}+3)}^2}dx$
Taking $4x^{-5}+2x^{-4}+3=z$
$\Rightarrow (-20x^{-6}-8x^{-5})dx=dz\Rightarrow f\left(z\right)=-\frac{1}{4}\int \frac{dz}{z^2}=\frac{1}{4z}+C\Rightarrow f\left(x\right)=\frac{x^5}{4(3x^5+2x+4)}+C$
As $f\left(0\right)=0\Rightarrow C=0$
So, $72\cdot f\left(1\right)=72\times \frac{1}{36}=2$