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

Sep 2, 2016

${x}^{2} / 2 - 2 \ln \left({x}^{2} + 4\right) + 2 a r c \tan \left(\frac{x}{2}\right) + C$.

#### Explanation:

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

Since the degree of poly. in Nr. $>$ that of in DR. , we have

to first perform long division. Instead, we proceed as under :

$N r . = {x}^{3} + 4$

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

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

$= \frac{x \left({x}^{2} + 4\right)}{\left({x}^{2} + 4\right)} - \frac{4 x}{{x}^{2} + 4} + \frac{4}{{x}^{2} + 4}$

$= x - \frac{4 x}{{x}^{2} + 4} + \frac{4}{{x}^{2} + 4}$

$\therefore I = \int \left[x - \frac{4 x}{{x}^{2} + 4} + \frac{4}{{x}^{2} + 4}\right] \mathrm{dx}$

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

$= {x}^{2} / 2 - 2 \int \frac{d \left({x}^{2} + 4\right)}{{x}^{2} + 4} + 4 \cdot \frac{1}{2} \cdot a r c \tan \left(\frac{x}{2}\right)$

$= {x}^{2} / 2 - 2 \ln \left({x}^{2} + 4\right) + 2 a r c \tan \left(\frac{x}{2}\right) + C$.

Enjoy maths.!