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

Question

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

Accepted Answer

Dividing numerator and denominator by $x^{14}$:
$$f(x)=\int\frac{5x^{-6}+7x^{-8}}{(2+x^{-7}+x^{-5})^2}dx$$

Let $2+x^{-7}+x^{-5}=t\Rightarrow(-7x^{-8}-5x^{-6})dx=dt$:
$$f(x)=\int\frac{-dt}{t^2}=t^{-1}+c=\frac1{2+x^{-7}+x^{-5}}+c$$

With $f(0)=0$: as $x\to0^+$, $x^{-7},x^{-5}\to\infty$ so the term $\to0$, giving $c=0$.

$$f(1)=\frac1{2+1+1}=\frac14$$

Solutions · 1