Question

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

Answer 1

Step 1. Simplify the given function.

In the integral of the function $f\left(x\right)={\int }_{}\frac{5x^8+7x^6}{{\left[x^2+1+2x^7\right]}^2}dx$ divide the numerator and denominator by $x^{14}$.

$\begin{array}{rcl}f\left(x\right)& =& {\int }_{}\frac{{\displaystyle \frac{5x^8+7x^6}{x^{14}}}}{{\left[{\displaystyle \frac{x^2+1+2x^7}{x^7}}\right]}^2}dx\\ & =& {\int }_{}\frac{{\displaystyle 5x^{-6}+7x^{-8}}}{{\left[{\displaystyle x^{-5}+x^{-1}+2}\right]}^2}dx\end{array}$

Now, let $\frac{1}{x^{-5}+x^{-7}+2}=t$, then $\frac{5x^{-6}+7x^8}{{\left[x^{-5}+x^{-1}+2\right]}^2}dx=dt$. So the integral becomes

$\begin{array}{rcl}{\int }_{}\frac{{\displaystyle 5x^{-6}+7x^{-8}}}{{\left[{\displaystyle x^{-5}+x^{-1}+2}\right]}^2}dx& =& {\int }_{}dt\\ & =& t+C\\ & =& \frac{1}{x^{-5}+x^{-7}+2}+C\left[t=\frac{1}{x^{-5}+x^{-7}+2}\right]\end{array}$

So the function is $f\left(x\right)=\begin{array}{rcl}& & \frac{1}{x^{-5}+x^{-7}+2}\end{array}+C$.

Step 2. Find the value of $C$.

The function $f\left(x\right)=\begin{array}{l}\frac{1}{x^{-5}+x^{-7}+2}\end{array}+C$ can be simplified as:

$f\left(x\right)=\begin{array}{l}\frac{x^7}{x^2+1+2x^7}\end{array}+C$

It is given that $f\left(0\right)=0$, so for $x=0$, the value of can be found as:

$$\begin{gathered} f\left(0\right)=\frac{0^7}{0^2+1+2\times 0^7}+C \\ \Rightarrow 0=0+C \\ \Rightarrow 0=C \end{gathered}$$

So, the function is $f\left(x\right)=\begin{array}{l}\frac{x^7}{x^2+1+2x^7}\end{array}$.

Step 3. Find the value of $k$.

For the function $f\left(x\right)=\begin{array}{l}\frac{x^7}{x^2+1+2x^7}\end{array}$, the value of $f\left(1\right)$ is given as:

$\begin{array}{rcl}f\left(1\right)& =& \begin{array}{l}\frac{1^7}{1^2+1+2\times 1^7}\end{array}\\ & =& \frac{1}{1+1+2}\\ & =& \frac{1}{4}\end{array}$

But it is given that $f\left(1\right)=\frac{1}{k}$, so the value of $k$ is $4$.

Hence, the value of $k$ is $4$.