Question

If $f\left( x \right)=\int{\dfrac{5{{x}^{8}}+7{{x}^{6}}}{{{\left( {{x}^{2}}+1+2{{x}^{7}} \right)}^{2}}}dx}$ and $f\left( 0 \right)=0$, then find the value of $f\left( 1 \right)$?
(a) $-\dfrac{1}{2}$,
(b) $\dfrac{1}{2}$,
(c) $-\dfrac{1}{4}$,
(d) $\dfrac{1}{4}$.

Answer 1

We start solving the problem by taking ${{x}^{7}}$ common inside the square of the denominator in integrand. We then make necessary arrangements to the integrand and we assume the $\dfrac{1}{{{x}^{5}}}+\dfrac{1}{{{x}^{7}}}+2=t$. We differentiate this and convert all the variables in integrand in terms of t. We then make the integration operations and replace t with original function. We then substitute 0 in place of x to get the value of constant of integration. We then substitute 1 in place of x and make subsequent calculations to get the required value.

Complete step by step answer:
According to the problem, we have given $f\left( x \right)=\int{\dfrac{5{{x}^{8}}+7{{x}^{6}}}{{{\left( {{x}^{2}}+1+2{{x}^{7}} \right)}^{2}}}dx}$. We need to find the value of $f\left( 1 \right)$ if given $f\left( 0 \right)=0$.
Let us first find the function $f\left( x \right)$ and then find the value of $f\left( 1 \right)$.
Let us consider $f\left( x \right)=\int{\dfrac{5{{x}^{8}}+7{{x}^{6}}}{{{\left( {{x}^{2}}+1+2{{x}^{7}} \right)}^{2}}}dx}$.
$\Rightarrow f\left( x \right)=\int{\dfrac{5{{x}^{8}}+7{{x}^{6}}}{{{\left( {{x}^{7}}\left( \dfrac{{{x}^{2}}}{{{x}^{7}}}+\dfrac{1}{{{x}^{7}}}+2 \right) \right)}^{2}}}dx}$.
$\Rightarrow f\left( x \right)=\int{\dfrac{5{{x}^{8}}+7{{x}^{6}}}{{{\left( {{x}^{7}} \right)}^{2}}{{\left( \dfrac{1}{{{x}^{5}}}+\dfrac{1}{{{x}^{7}}}+2 \right)}^{2}}}dx}$.
$\Rightarrow f\left( x \right)=\int{\dfrac{5{{x}^{8}}+7{{x}^{6}}}{\left( {{x}^{14}} \right){{\left( \dfrac{1}{{{x}^{5}}}+\dfrac{1}{{{x}^{7}}}+2 \right)}^{2}}}dx}$.
$\Rightarrow f\left( x \right)=\int{\dfrac{\dfrac{5{{x}^{8}}}{{{x}^{14}}}+\dfrac{7{{x}^{6}}}{{{x}^{14}}}}{{{\left( \dfrac{1}{{{x}^{5}}}+\dfrac{1}{{{x}^{7}}}+2 \right)}^{2}}}dx}$.
$\Rightarrow f\left( x \right)=\int{\dfrac{\dfrac{5}{{{x}^{6}}}+\dfrac{7}{{{x}^{8}}}}{{{\left( \dfrac{1}{{{x}^{5}}}+\dfrac{1}{{{x}^{7}}}+2 \right)}^{2}}}dx}$ ---(1).
Let us assume $\dfrac{1}{{{x}^{5}}}+\dfrac{1}{{{x}^{7}}}+2=t$ ---(2).
We differentiate this on both sides.
$\Rightarrow d\left( \dfrac{1}{{{x}^{5}}}+\dfrac{1}{{{x}^{7}}}+2 \right)=dt$.
$\Rightarrow d\left( \dfrac{1}{{{x}^{5}}} \right)+d\left( \dfrac{1}{{{x}^{7}}} \right)+d\left( 2 \right)=dt$.
We know that $d\left( \dfrac{1}{{{x}^{n}}} \right)=\dfrac{-n}{{{x}^{n+1}}}dx$ and $d\left( a \right)=0$.
$\Rightarrow \left( \dfrac{-5}{{{x}^{6}}} \right)dx+\left( \dfrac{-7}{{{x}^{8}}} \right)dx+0=dt$.
$\Rightarrow \left( \dfrac{-5}{{{x}^{6}}}+\left( \dfrac{-7}{{{x}^{8}}} \right) \right)dx=dt$.
$\Rightarrow -\left( \dfrac{5}{{{x}^{6}}}+\dfrac{7}{{{x}^{8}}} \right)dx=dt$.
$\Rightarrow \left( \dfrac{5}{{{x}^{6}}}+\dfrac{7}{{{x}^{8}}} \right)dx=-dt$ ---(3).
Now, we substitute equation (2) and (3) in equation (1).
$\Rightarrow f\left( x \right)=\int{\dfrac{1}{{{\left( t \right)}^{2}}}\left( -dt \right)}$.
$\Rightarrow f\left( x \right)=-\int{\dfrac{1}{{{\left( t \right)}^{2}}}dt}$.
$\Rightarrow f\left( x \right)=-\int{{{t}^{-2}}dt}$.
We know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$.
$\Rightarrow f\left( x \right)=-\left( \dfrac{{{t}^{-2+1}}}{-2+1} \right)+C$.
$\Rightarrow f\left( x \right)=-\left( \dfrac{{{t}^{-1}}}{-1} \right)+C$.
$\Rightarrow f\left( x \right)=\dfrac{1}{t}+C$ ---(4).
From equation (2) we have $t=\dfrac{1}{{{x}^{5}}}+\dfrac{1}{{{x}^{7}}}+2$. We use this is equation (4).
$\Rightarrow f\left( x \right)=\dfrac{1}{\left( \dfrac{1}{{{x}^{5}}}+\dfrac{1}{{{x}^{7}}}+2 \right)}+C$.
$\Rightarrow f\left( x \right)=\dfrac{1}{\left( \dfrac{1+{{x}^{2}}+2{{x}^{7}}}{{{x}^{7}}} \right)}+C$.
$\Rightarrow f\left( x \right)=\dfrac{{{x}^{7}}}{\left( 1+{{x}^{2}}+2{{x}^{7}} \right)}+C$ ---(5), where C is constant of integration.
According to the problem, we have $f\left( 0 \right)=0$. So, let us $x=0$ in equation (5).
$\Rightarrow f\left( 0 \right)=\dfrac{{{0}^{7}}}{\left( 1+{{0}^{2}}+2\left( {{0}^{7}} \right) \right)}+C$.
$\Rightarrow 0=\dfrac{0}{\left( 1+0+2\left( 0 \right) \right)}+C$.
$\Rightarrow 0=0+C$.
$\Rightarrow 0=C$.
Let us substitute the value of C in equation (5).
$\Rightarrow f\left( x \right)=\dfrac{{{x}^{7}}}{\left( 1+{{x}^{2}}+2{{x}^{7}} \right)}+0$.
$\Rightarrow f\left( x \right)=\dfrac{{{x}^{7}}}{\left( 1+{{x}^{2}}+2{{x}^{7}} \right)}$ ---(6).
Let us substitute $x=1$ in equation (6) to get the value of $f\left( 1 \right)$.
$\Rightarrow f\left( 1 \right)=\dfrac{{{1}^{7}}}{\left( 1+{{1}^{2}}+2\left( {{1}^{7}} \right) \right)}$.
$\Rightarrow f\left( 1 \right)=\dfrac{1}{\left( 1+1+2\left( 1 \right) \right)}$.
$\Rightarrow f\left( 1 \right)=\dfrac{1}{\left( 2+2 \right)}$.
$\Rightarrow f\left( 1 \right)=\dfrac{1}{4}$.
We have found the value of $f\left( 1 \right)$ as $\dfrac{1}{4}$.
∴ The value of $f\left( 1 \right)$ is $\dfrac{1}{4}$.

So, the correct answer is “Option D”.

Note: We should not forget to add constant of integration as we are doing indefinite integral. We should make sure that the function obtained should in terms of x as the value of the function given in the problem is defined by substituting in place of x. We should not make mistakes while finding the integration and substitutions. We can also solve this problem by taking partial fractions but it will take a heavy time to make calculations.