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JEE MAIN 2019
09-01-19 S2
Question
If $f(x)=\int \frac{5 x^8+7 x^6}{\left(x^2+1+2 x\right)^2} d x,(x \geq 0)$, and $f(0)=0$, then the value of $f(1)$ is
Select the correct option:
A
$\frac{1}{2}$
B
$\frac{1}{4}$
C
$-\frac{1}{2}$
D
$-\frac{1}{4}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & f(x)=\int \frac{5 x^8+7 x^6}{\left(x^2+1+2 x\right)^2} d x, x \geq 0 \\ & =\int \frac{5 x^8+7 x^6}{x^{14}\left(x^{-5}+x^7+2\right)^2} d x \\ & =\int \frac{5 x^{-6}+7 x^{-8}}{\left(2+x^{-7}+x^{-5}\right)^2} d x \\ & \text { Let } 2+x^{-7}+x^{-5}=t \\ & \quad\left(-7 x^{-8}-5 x^{-6}\right) d x=d t \\ & f(x)=\int \frac{-d t}{t^2}=\int-t^{-2} d t=t^{-1}+c \\ & f(x)=\frac{1}{2+x^{-7}+x^{-5}}+c, f(0)=0 \Rightarrow c=0 \\ & \therefore f(1)=\frac{1}{4}\end{aligned}$
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