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Question: Answered & Verified by Expert
If $f(x)=\int \frac{5 x^8+7 x^6}{\left(x^2+1+2 x^7\right)^2} d x, x \geq 0$ and $f(0)=0$, then value of $f(1)$ is
MathematicsIndefinite IntegrationMHT CETMHT CET 2022 (06 Aug Shift 1)
Options:
  • A $-\frac{1}{2}$
  • B $\frac{1}{4}$
  • C $-\frac{1}{4}$
  • D $\frac{1}{2}$
Solution:
2636 Upvotes Verified Answer
The correct answer is: $\frac{1}{4}$
$\begin{aligned} & f(x)=\int \frac{5 x^8+7 x^6}{\left(x^2+1+2 x^7\right)^2} d x \\ & =\int \frac{5 x^8+7 x^6}{x^{14}\left(\frac{1}{x^5}+\frac{1}{x^7}+2\right)^2} d x \\ & =\int \frac{5 x^{-6}+7 x^{-8}}{\left(\frac{1}{x^5}+\frac{1}{x^7}+2\right)^2} d x \\ & =-\int \frac{d t}{t^2}\left[\text { where } t=\frac{1}{x^5}+\frac{1}{x^7}+2\right] \\ & =\frac{1}{t}+C\end{aligned}$
$\begin{aligned} & =\frac{1}{\left(\frac{1}{x^5}+\frac{1}{x^7}+2\right)}+C \\ & \Rightarrow f(x)=\frac{x^7}{x^2+1+2 x^7}+C \\ & \because f(0)=0 \Rightarrow c=0 \text { i.e. } f(x)=\frac{x^7}{x^2+1+2 x^7} \\ & \Rightarrow f(1)=\frac{1}{4}\end{aligned}$

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