If f(x) = (5x^8+7x^6)/((x^2+1+2x^7)^2),dx, x0, f(0)=0 and f(1)=1K, then the value of K is …

Question

If $f(x) = \displaystyle\int\frac{5x^8+7x^6}{(x^2+1+2x^7)^2}\,dx$, $x\ge0$, $f(0)=0$ and $f(1)=\dfrac1K$, then the value of $K$ is

Answer: 4

Accepted Answer

Divide numerator and denominator by $x^{14}$: $I = \displaystyle\int\frac{5x^{-6}+7x^{-8}}{(x^{-5}+x^{-7}+2)^2}\,dx$. Let $z=x^{-5}+x^{-7}+2$, $dz=-(5x^{-6}+7x^{-8})\,dx$:

$$I = -\int\frac{dz}{z^2} = \frac1z+c = \frac1{x^{-5}+x^{-7}+2}+c = \frac{x^7}{x^2+1+2x^7}+c$$

Using $f(0)=0$: $c=0$. So $f(1)=\dfrac1{1+1+2}=\dfrac14$, giving $K=4$.

Solutions · 1