Question

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

Answer 1

Hint: First we divide the numerator and denominator by ${x^{14}}$. By using the substitution method we will calculate the value of $f\left( x \right)$. After that, substitute $x=1$ in the equation $f\left( x \right)$ and compare with $f\left( 1 \right) = \dfrac{1}{k}$ to get the value of $k$.

Formula Used:
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
$\int {\dfrac{1}{{{x^2}}}} dx = - \dfrac{1}{x} + c$

Complete step by step solution:
Given integration is
$f\left( x \right) = \int {\dfrac{{5{x^8} + 7{x^6}}}{{{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}dx} $
Taking common ${x^7}$ from the numerator
$f\left( x \right) = \int {\dfrac{{5{x^8} + 7{x^6}}}{{{{\left\{ {{x^7}\left( {\dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^7}}} + 2} \right)} \right\}}^2}}}dx} $
$ \Rightarrow f\left( x \right) = \int {\dfrac{{5{x^8} + 7{x^6}}}{{{x^{14}}{{\left( {\dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^7}}} + 2} \right)}^2}}}dx} $
Now divide the numerator and denominator by ${x^{14}}$
$f\left( x \right) = \int {\dfrac{{\dfrac{5}{{{x^6}}} + \dfrac{7}{{{x^8}}}}}{{{{\left( {\dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^7}}} + 2} \right)}^2}}}dx} $
Assume that $\dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^7}}} + 2 = z$
Differentiate both sides
$\left( {\dfrac{{ - 5}}{{{x^6}}} + \dfrac{{ - 7}}{{{x^8}}}} \right)dx = dz$
$ \Rightarrow \left( {\dfrac{5}{{{x^6}}} + \dfrac{7}{{{x^8}}}} \right)dx = - dz$
Putting $\left( {\dfrac{5}{{{x^6}}} + \dfrac{7}{{{x^8}}}} \right)dx = - dz$ and $\dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^7}}} + 2 = z$ in the given function $f\left( x \right) = \int {\dfrac{{\dfrac{5}{{{x^6}}} + \dfrac{7}{{{x^8}}}}}{{{{\left( {\dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^7}}} + 2} \right)}^2}}}dx} $.
$ \Rightarrow f\left( x \right) = \int {\dfrac{1}{{{{\left( z \right)}^2}}}\left( { - dz} \right)} $
Now applying the integration formula $\int {\dfrac{1}{{{x^2}}}} dx = - \dfrac{1}{x} + c$
$ \Rightarrow f\left( x \right) = \dfrac{1}{z} + C$
Now substitute the value of $z$
$ \Rightarrow f\left( x \right) = \dfrac{1}{{\dfrac{1}{{{x^5}}} + \dfrac{1}{{{x^7}}} + 2}} + C$
Simplify the above equation
$ \Rightarrow f\left( x \right) = \dfrac{{{x^7}}}{{{x^2} + 1 + 2{x^7}}} + C$
Now putting $x = 0$ to calculate the value of $C$
$ \Rightarrow f\left( 0 \right) = \dfrac{{{0^7}}}{{{0^2} + 1 + 2 \cdot {0^7}}} + C$
Now putting the value of $f\left( 0 \right)$
$ \Rightarrow 0 = \dfrac{{{0^7}}}{{{0^2} + 1 + 2 \cdot {0^7}}} + C$
$ \Rightarrow 0 = C$
Therefore, the given function becomes $f\left( x \right) = \dfrac{{{x^7}}}{{{x^2} + 1 + 2{x^7}}}$.
Now putting $x = 1$ in $f\left( x \right) = \dfrac{{{x^7}}}{{{x^2} + 1 + 2{x^7}}}$.
$f\left( 1 \right) = \dfrac{{{1^7}}}{{{1^2} + 1 + 2 \cdot {1^7}}}$
Simplify the right-hand side expression
$f\left( 1 \right) = \dfrac{1}{4}$
Given that $f\left( 1 \right) = \dfrac{1}{k}$.
Thus, $\dfrac{1}{4} = \dfrac{1}{k}$
Which implies $k = 4$.

Note: If an integration is not solved directly, then we will apply the substitution method. In the given question, we are unable to solve it directly, so apply the substitution method. Then substitute x=1 in the solution and calculate the value of k.