Evaluate the integral:


Evaluate the integral:

$\int \frac{x^{3}}{x^{4}+x^{2}+1} d x$


Let, $I=\int \frac{x^{3}}{x^{4}+x^{2}+1} d x$

$I=\int \frac{x^{2} x}{\left(x^{2}\right)^{2}+x^{2}+1} d x$

If we assume $x^{2}$ to be an another variable, we can simplify the integral as derivative of $x^{2}$ i.e. $x$ is present in numerator.

Let, $x^{2}=u$

$\Rightarrow 2 \mathrm{x} \mathrm{dx}=\mathrm{du}$

$\Rightarrow \mathrm{x} \mathrm{dx}=1 / 2 \mathrm{du}$

$\therefore I=\frac{1}{2} \int \frac{u}{u^{2}+u+1} d u$

As, $\frac{\mathrm{d}}{\mathrm{du}}\left(\mathrm{u}^{2}+\mathrm{u}+1\right)=2 \mathrm{u}+1$

$\therefore$ Let, $u=A(2 u+1)+B$

$\Rightarrow u=2 A u+A+B$

On comparing both sides -

We have,

$2 \mathrm{~A}=1 \Rightarrow \mathrm{A}=1 / 2$

$A+B=0 \Rightarrow B=-A=-1 / 2$


$I=\frac{1}{2} \int \frac{\frac{1}{2}(2 u+1) \frac{1}{2}}{u^{2}+u+1} d u$

$\therefore I=\frac{1}{4} \int \frac{(2 \mathrm{u}+1)}{\mathrm{u}^{2}+\mathrm{u}+1} \mathrm{du}+\frac{1}{2} \int \frac{-\frac{1}{2}}{\mathrm{u}^{2}+\mathrm{u}+1} \mathrm{du}$

Let, $I_{1}=\frac{1}{4} \int \frac{(2 u+1)}{u^{2}+u+1} d u$ and $I_{2}=-\frac{1}{4} \int \frac{1}{u^{2}+u+1} d u$

Now, $I=I_{1}+I_{2} \ldots$ eqn 1

We will solve $\mathrm{I}_{1}$ and $\mathrm{I}_{2}$ individually.

$\mathrm{As}, \mathrm{I}_{1}=\frac{1}{4} \int \frac{(2 \mathrm{u}+1)}{\mathrm{u}^{2}+\mathrm{u}+1} \mathrm{du}$

Let $v=u^{2}+u+1 \Rightarrow d v=(2 u+1) d u$

$\therefore \mathrm{I}_{1}$ reduces to $\frac{1}{4} \int \frac{\mathrm{dv}}{\mathrm{v}}$


$\mathrm{I}_{1}=\frac{1}{4} \int \frac{\mathrm{dv}}{\mathrm{v}}=\frac{1}{4} \log |\mathrm{v}|+\mathrm{C}\left\{\because \int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|+\mathrm{C}\right\}$

On substituting value of $u$, we have:

$\mathrm{I}_{1}=\frac{1}{4} \log \left|\mathrm{u}^{2}+\mathrm{u}+1\right|+\mathrm{C} \ldots . . \mathrm{eqn} 2$

As, $I_{2}=-\frac{1}{4} \int \frac{1}{u^{2}+u+1}$ du and we don't have any derivative of function present in denominator. $\therefore$ we will use some special integrals to solve the problem.

As denominator doesn't have any square root term. So one of the following two integrals will solve the problem.

i) $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$ ii $\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$

Now we have to reduce $\mathrm{I}_{2}$ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of $x$ is seen in denominator.

$\therefore I_{2}=-\frac{1}{4} \int \frac{1}{u^{2}+u+1} d u$

$\Rightarrow I_{2}=-\frac{1}{4} \int \frac{1}{\left\{u^{2}+2\left(\frac{1}{2}\right) u+\left(\frac{1}{2}\right)^{2}\right\}+1-\left(\frac{1}{2}\right)^{2}} d u$

Using: $a^{2}+2 a b+b^{2}=(a+b)^{2}$

We have:

$\mathrm{I}_{2}=-\frac{1}{4} \int \frac{1}{\left(\mathrm{u}+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{2}}{2}\right)^{2}} \mathrm{du}$

$\therefore I_{2}=-\frac{1}{4} \frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{u+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$

$\therefore I_{2}=-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 u+1}{\sqrt{3}}\right)+C \ldots \operatorname{eqn} 3$

From eqn 1, we have:


Using eqn 2 and 3 , we get -

$\mathrm{I}=\frac{1}{4} \log \left|\mathrm{u}^{2}+\mathrm{u}+1\right|-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \mathrm{u}+1}{\sqrt{3}}\right)+\mathrm{C}$

Putting value of $u$ in I:]

$I=\frac{1}{4} \log \left|x^{2^{2}}+x^{2}+1\right|-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+C$

$I=\frac{1}{4} \log \left|x^{4}+x^{2}+1\right|-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}+1}{\sqrt{3}}\right)+C$

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