# How do you use partial fractions to find the integral int (x^2+x+3)/(x^4+6x^2+9)dx? Jan 18, 2017

$\frac{1}{\sqrt{3}} {\tan}^{- 1} \left(\frac{x}{\sqrt{3}}\right) - \frac{1}{2 \left({x}^{2} + 3\right)} + C$

#### Explanation:

Noticing that the denominator is a perfect square, we see there is no need to use partial fractions.

$\frac{{x}^{2} + x + 3}{{x}^{4} + 6 {x}^{2} + 9} = \frac{{x}^{2} + 3 + x}{{x}^{2} + 3} ^ 2 = \frac{\cancel{{x}^{2} + 3}}{{x}^{2} + 3} ^ \cancel{2} + \frac{x}{{x}^{2} + 3} ^ 2$
$= \frac{1}{{x}^{2} + 3} + \frac{x}{{x}^{2} + 3} ^ 2$
The first term is a standard integral of the form $\frac{1}{{x}^{2} + {a}^{2}}$ with $a = \sqrt{3}$, yielding $\left(\frac{1}{\sqrt{3}}\right) {\tan}^{- 1} \left(\frac{x}{\sqrt{3}}\right)$.

The second term is of the form where the the numerator is the derivative of the contents of the brackets in the denominator, save for a factor of $2$, and so can be integrated at sight. (Alternatively substitute $u = {x}^{2} + 3$, $\frac{\mathrm{dx}}{\mathrm{du}} = \frac{1}{2 x}$ to give $\frac{1}{2} \int {u}^{- 2} \mathrm{du}$.)

In general, $\frac{d}{\mathrm{dx}} \left(\frac{1}{f \left(x\right)}\right) = - \frac{f p r i m e \left(x\right)}{f \left(x\right)} ^ 2$ so $\int \frac{f p r i m e \left(x\right)}{f \left(x\right)} ^ 2 \mathrm{dx} = - \frac{1}{f \left(x\right)}$. In this case, $f \left(x\right) = {x}^{2} + 3$ and $f p r i m e \left(x\right) = 2 x$.