# How do you evaluate the integral int (x^4+3x+1)/(x^2+x+1)dx?

May 22, 2017

${x}^{3} / 3 - {x}^{2} / 2 + 2 \ln \left({x}^{2} + x + 1\right) - \frac{2}{\sqrt{3}} a r c \tan \left(\frac{2 x + 1}{\sqrt{3}}\right) + C .$

#### Explanation:

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

We notice that the given Integrand is an Improper Rational

Function, (i.e., the degree of the Nr. poly. is more than that of the

Dr. poly.); so, our first task is to make it Proper. This is usually

done by Long Division, but we proceed as under :

$\because , {x}^{4} + 3 x + 1 = {x}^{2} \left({x}^{2} + x + 1\right) - x \left({x}^{2} + x + 1\right) + 4 x + 1 ,$

$\therefore \frac{{x}^{4} + 3 x + 1}{{x}^{2} + x + 1} = {x}^{2} - x + \frac{4 x + 1}{{x}^{2} + x + 1} .$

$\Rightarrow I = \int \left\{{x}^{2} - x + \frac{4 x + 1}{{x}^{2} + x + 1}\right\} \mathrm{dx} ,$

$= {x}^{3} / 3 - {x}^{2} / 2 + \int \frac{4 x + 1}{{x}^{2} + x + 1} \mathrm{dx} .$

Here, $\frac{d}{\mathrm{dx}} \left({x}^{2} + x + 1\right) = 2 x + 1 , \text{ so, we take } 4 x + 1 = 2 \left(2 x + 1\right) - 1 ,$ and get,

$I = {x}^{3} / 3 - {x}^{2} / 2 + \int \frac{2 \left(2 x + 1\right) - 1}{{x}^{2} + x + 1} \mathrm{dx} ,$

$= {x}^{3} / 3 - {x}^{2} / 2 + 2 \int \frac{\frac{d}{\mathrm{dx}} \left({x}^{2} + x + 1\right)}{{x}^{2} + x + 1} \mathrm{dx} - \int \frac{1}{{x}^{2} + x + 1} \mathrm{dx} ,$

$= {x}^{3} / 3 - {x}^{2} / 2 + 2 \ln \left({x}^{2} + x + 1\right) - \int \frac{1}{\left({x}^{2} + x + \frac{1}{4}\right) + \frac{3}{4}} \mathrm{dx} ,$

=x^3/3-x^2/2+2ln(x^2+x+1)-int1/{(x+1/2)^2+(sqrt3/2)^2dx,

$= {x}^{3} / 3 - {x}^{2} / 2 + 2 \ln \left({x}^{2} + x + 1\right) - \frac{1}{\frac{\sqrt{3}}{2}} a r c \tan \left\{\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right\} ,$

$\therefore I = {x}^{3} / 3 - {x}^{2} / 2 + 2 \ln \left({x}^{2} + x + 1\right) - \frac{2}{\sqrt{3}} a r c \tan \left(\frac{2 x + 1}{\sqrt{3}}\right) + C .$

Enjoy Maths.!