How do you find int (x^3+x^2+2x+1)/((x^2+1)(x^2+2)) dx using partial fractions?

Mar 19, 2018

$\frac{1}{2} \ln | {x}^{2} + 1 | + \frac{1}{\sqrt{2}} {\tan}^{-} 1 \left(\frac{x}{\sqrt{2}}\right) + c$

Explanation:

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

Rearranging the terms in numerator,

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

Simplifying according to required factors in denominator,

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

$= \int \frac{x \cancel{\left({x}^{2} + 2\right)}}{\left({x}^{2} + 1\right) \cancel{\left({x}^{2} + 2\right)}} \mathrm{dx} + \int \frac{\cancel{\left({x}^{2} + 1\right)}}{\cancel{\left({x}^{2} + 1\right)} \left({x}^{2} + 2\right)} \mathrm{dx}$

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

$= \frac{1}{2} \int \frac{2 x}{{x}^{2} + 1} \mathrm{dx} + \int \frac{1}{{x}^{2} + {\left(\sqrt{2}\right)}^{2}} \mathrm{dx}$

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

$= \frac{1}{2} \ln | {x}^{2} + 1 | + \frac{1}{\sqrt{2}} {\tan}^{-} 1 \left(\frac{x}{\sqrt{2}}\right) + c$