Całka do policzenia

[moczul]

\(\displaystyle{ \int\limits_{0}^{+\infty}4te^{-2t}dt}\)

[luka52]

\(\displaystyle{ I = 4 t t e^{-2t} \, \mbox{d}t}\)
Przez części:
\(\displaystyle{ u = 4t, \quad \mbox{d}v = e^{-2t} \mbox{d}t\\
\mbox{d}u = 4 \mbox{d}t, \quad v = - \frac{1}{2}e^{-2t}\\
I = -2t e^{-2t} + 2 t e^{-2t} \, \mbox{d}t = -2 t e^{-2t} - e^{-2t} + C}\)
Zatem:
\(\displaystyle{ \int \limits_0^{+ } 4 t e^{-2t} \, \mbox{dt} = \lim_{x \to +\infty} \left( -2 x e^{-2x} - e^{-2x} \right) - (-1) = 1}\)

[moczul]

Jeszcze dwi calki z ktorymi mam klopot

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


\(\displaystyle{ \int\limits_{2}^{+\infty}\frac{dx}{x(x+2)}dt}\)

[LecHu :)]

1)\(\displaystyle{ \frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}}\)
\(\displaystyle{ ...=ln(\lim\limits_{x\to\infty}\frac{x}{x+1})-ln\frac{2}{3}=ln\frac{3}{2}}\)

[luka52]

ad 2)
\(\displaystyle{ = \frac{1}{2} \int\limits_{2}^{+\infty} \left( \frac{1}{x} - \frac{1}{2+x} \right) \, = \frac{1}{2} \left[ \ln \left| \frac{x}{x+2} \right| \right]_2^{+\infty} = \frac{\ln 2}{2}}\)