# How do you integrate int (x^3+x^2+x+2)/(x^4+x^2) using partial fractions?

## How do you integrate $\int \frac{{x}^{3} + {x}^{2} + x + 2}{{x}^{4} + {x}^{2}} \mathrm{dx}$ using partial fractions?

Aug 28, 2016

$\ln | x | - \frac{2}{x} - a r c \tan x + C$

#### Explanation:

Let $I = \int \frac{{x}^{3} + {x}^{2} + x + 2}{{x}^{4} + {x}^{2}} \mathrm{dx} = \int \frac{{x}^{3} + {x}^{2} + x + 2}{{x}^{2} \left({x}^{2} + 1\right)} \mathrm{dx}$

On the first hand inspection, we find that we have no go but to use

the Method of Partial Fraction to decompose the Integrand, but,

$\text{The Nr.}$

$= \left({x}^{3} + {x}^{2} + x + 2\right) = \underline{{x}^{3} + x} + \underline{{x}^{2} + 1} + 1$

$= x \left({x}^{2} + 1\right) + 1 \left({x}^{2} + 1\right) + 1 = \left({x}^{2} + 1\right) \left(x + 1\right) + 1$.

Hence, $\frac{{x}^{3} + {x}^{2} + x + 2}{{x}^{2} \left({x}^{2} + 1\right)} = \frac{\left({x}^{2} + 1\right) \left(x + 1\right) + 1}{{x}^{2} \left({x}^{2} + 1\right)}$

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

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

=(x+1)/x^2+{(x^2+1)-x^2)/(x^2(x^2+1)

=x/x^2+1/x^2+(x^2+1)/(x^2(x^2+1))-x^2/(x^2(x^2+1)

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

Therefore,

$I = \int \left[\frac{1}{x} + \frac{2}{x} ^ 2 - \frac{1}{{x}^{2} + 1}\right] \mathrm{dx}$

$= \ln | x | - \frac{2}{x} - a r c \tan x + C$

Enjoy maths.!