Question

The integral $\int{\dfrac{3{{x}^{13}}+2{{x}^{11}}}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{4}}}dx}$ is equal to: (where C is a constant of integration)
(a) $\dfrac{{{x}^{4}}}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{3}}}+C$
(b) $\dfrac{{{x}^{12}}}{6{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{3}}}+C$
(c) $\dfrac{{{x}^{4}}}{6{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{3}}}+C$
(d) $\dfrac{{{x}^{12}}}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{3}}}+C$

Answer 1

First, take the highest power of x, ${{x}^{16}}$ from numerator and denominator. Then observe the remaining integral, try that numerator becomes the differentiation of the denominator terms, with some manipulations. Assume the denominator term whose differentiation is in the numerator as another variable and integrate with respect to the new variable.

Complete step-by-step answer:
We are given an integrand which seems complicated to integrate at first , so we will first try to simplify it .
To simplify it first take common the highest power of $x$ , ${{x}^{^{16}}}$ from numerator and denominator,
So, first consider the given integral,
$\int{\dfrac{3{{x}^{13}}+2{{x}^{11}}}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{4}}}dx}$
Now, take common ${{x}^{^{16}}}$ from numerator and denominator, and get,
$\begin{aligned} & \int{\dfrac{\left( 3{{x}^{13}}+2{{x}^{11}} \right)}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{4}}}dx} \\\ & =\int{\dfrac{{{x}^{16}}\left( \dfrac{3}{{{x}^{3}}}+\dfrac{2}{{{x}^{5}}} \right)}{{{x}^{16}}{{\left( 2+\dfrac{3}{{{x}^{2}}}+\dfrac{4}{{{x}^{4}}} \right)}^{4}}}dx} \\\ & =\int{\dfrac{\left( \dfrac{3}{{{x}^{3}}}+\dfrac{2}{{{x}^{5}}} \right)}{{{\left( 2+\dfrac{3}{{{x}^{2}}}+\dfrac{4}{{{x}^{4}}} \right)}^{4}}}dx} \end{aligned}$
Now, we have little simplified integrand,
Now, observe that the term under the power 4 has the differentiation similar to the terms in numerator,
So, Let,
$2+\dfrac{3}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}}=t$
Now, differentiating on both the sides, we get,
$\begin{aligned} & \left( 0-\dfrac{6}{{{x}^{3}}}-\dfrac{4}{{{x}^{5}}} \right)dx=dt \\\ & \Rightarrow \left( \dfrac{6}{{{x}^{3}}}-\dfrac{4}{{{x}^{5}}} \right)dx=dt \\\ & \Rightarrow -2\left( \dfrac{3}{{{x}^{3}}}+\dfrac{2}{{{x}^{5}}} \right)dx=dt \\\ & \Rightarrow \left( \dfrac{3}{{{x}^{3}}}+\dfrac{2}{{{x}^{5}}} \right)dx=-\dfrac{dt}{2} \\\ \end{aligned}$
Now, converting the integral in terms of $t$ , we get,
$\begin{aligned} & =\int{\dfrac{-\dfrac{1}{2}}{{{t}^{4}}}dt} \\\ & =-\dfrac{1}{2}\int{{{t}^{-4}}dt} \\\ & =-\dfrac{1}{2}\dfrac{{{t}^{-3}}}{-3}+C \\\ & =\dfrac{1}{6}{{t}^{-3}}+C \\\ \end{aligned}$
Now, outing the value of t, we get,
$\dfrac{{{x}^{12}}}{6{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{3}}}+C$

$\begin{aligned} & =\dfrac{1}{6{{\left( 2+\dfrac{3}{{{x}^{2}}}+\dfrac{1}{{{x}^{4}}} \right)}^{3}}}+C \\\ & =\dfrac{1}{\dfrac{6{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{3}}}{{{x}^{12}}}}+C \\\ & =\dfrac{{{x}^{12}}}{6{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{3}}}+C \\\ \end{aligned}$

Hence, the value of the integral $\int{\dfrac{3{{x}^{13}}+2{{x}^{11}}}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{4}}}dx}$ = $\dfrac{{{x}^{12}}}{6{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{3}}}+C$

So, the correct answer is “Option b”.

Note: There is an alternate method of this question through observing the options of the questions.
See the options (a) and (c) they seem similar, difference is just because of constant terms. Same is the case with options (b) and (d). First, let the integral value is equal to $\dfrac{A{{x}^{4}}}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{3}}}+C$, so if this is the integral value then its differentiation will be equal to $\dfrac{3{{x}^{13}}+2{{x}^{11}}}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{4}}}$ . But, when you’ll differentiate it, it is not going to seem like this , $\dfrac{3{{x}^{13}}+2{{x}^{11}}}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{4}}}$,hence, absurd condition . So, now let that the integral value be equal to $\dfrac{B{{x}^{12}}}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{3}}}+C$. Now, when you’ll differentiate it is going to seem like the $\dfrac{3{{x}^{13}}+2{{x}^{11}}}{{{\left( 2{{x}^{4}}+3{{x}^{2}}+1 \right)}^{4}}}$. Equate with this and hence find the value of $B$ .