obliczyć całkę

[jeremi18]

\(\displaystyle{ \int\limits_{1}^{\infty} \frac{dx}{x^{2}-4x+5}}\)

[Jestemfajny]

Zamien to tak:
\(\displaystyle{ \int\limits_{1}^{\infty} \frac{dx}{(x-2)^{2}+1}}\) i podstaw \(\displaystyle{ t=x-2 \ \ dt=dx
}\)

\(\displaystyle{ \int\limits_{1}^{\infty} \frac{dt}{t^{2}+1}}\)

[scyth]

\(\displaystyle{ \int\limits_{1}^{\infty} \frac{dx}{x^{2}-4x+5} = \int\limits_{1}^{\infty} \frac{dx}{(x-2)^2+1}= \left[\arctan{(x-2)} \right]_{1}^{\infty}=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}}\)