Question:
If $f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0), f(0)=0 \mathrm{v}$ and $f(1)=\frac{1}{\mathrm{~K}}$, then the value of $\mathrm{K}$ is
Solution:
$f(x)=\int \frac{\left(5 x^{8}+7 x^{6}\right) d x}{x^{14}\left(x^{-5}+x^{-7}+2\right)^{2}}$
Let $x^{-5}+x^{-7}+2=t$
$\left(-5 x^{-6}-7 x^{-8}\right) d x=d t$
$\Rightarrow f(x)=\int-\frac{d t}{t^{2}}=\frac{1}{t}+c$
$f(x)=\frac{x^{7}}{x^{2}+1+2 x^{7}}$
$f(1)=\frac{1}{4}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.