# How do you test the improper integral int (x(1+x^2)^-2)dx from [0,oo) and evaluate if possible?

Mar 17, 2017

${\int}_{0}^{\infty} x {\left(1 + {x}^{2}\right)}^{-} 2 \mathrm{dx} = \frac{1}{2}$

#### Explanation:

Evaluate first the indefinite integral:

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

by substituting $\left(1 + {x}^{2}\right) = t$, so that $\mathrm{dt} = 2 x \mathrm{dx}$:

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

Now we have:

$F \left(0\right) = - \frac{1}{2} + C$

${\lim}_{x \to \infty} F \left(x\right) = C$

So the indefinite integral is convergent and we have:

${\int}_{0}^{\infty} x {\left(1 + {x}^{2}\right)}^{-} 2 \mathrm{dx} = \left({\lim}_{x \to \infty} F \left(x\right)\right) - F \left(0\right) = \frac{1}{2}$