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

Jan 11, 2017

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

#### Explanation:

Simplify the integral using partial fractions:

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

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

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

{ color(white)([)color(black)((A+B=0), (2A+B =1))]

{ color(white)([)color(black)((A= -B), (2A+B =1))]

{ color(white)([)color(black)((A= -B), (2A-A =1))]

{ color(white)([)color(black)((A= 1), (B=-1))]color(black)

So:

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

Let's rewrite the second addendum as:

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

and we have:

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