Question

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

• A. $-\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $-\frac{1}{4}$

Answer 1

The correct option is D $-\frac{1}{4}$

$f\left(x\right)=\int \frac{5x^8+7x^6}{(x^2+1+2x^7{)}^2}dx$

$=\int \frac{5x^{-6}+7x^{-8}}{(x^{-5}+x^{-7}+2{)}^2}dx$

Let $x^{-5}+x^{-7}+2=t$
$\Rightarrow (-5x^{-6}-7x^{-8})dx=dt$

$\therefore f\left(x\right)=-\int \frac{1}{t^2}dt$
$\Rightarrow f\left(x\right)=\frac{1}{t}+C$
$\Rightarrow f\left(x\right)=\frac{1}{x^{-5}+x^{-7}+2}+C$
$\therefore f\left(0\right)=\frac{1}{2}+C$
$\Rightarrow 0=\frac{1}{2}+C$
$\Rightarrow C=-\frac{1}{2}$
Thus, $f\left(x\right)=\frac{1}{x^{-5}+x^{-7}+2}-\frac{1}{2}$
Hence, $f\left(1\right)=\frac{1}{(1+1+2)}-\frac{1}{2}$
$=-\frac{1}{4}$