Matematické Fórum

nubijska princess · 08. 01. 2008 14:57

pomuze mi prosim nekdo s temito integraly? diky

int(sqrt ln^3x)/x

int(x-3)/x^2 - 2x - 5

rozlozte na parcialni zlomky:
4x - 8/x^2(x^2 - 4)

plisna · 08. 01. 2008 15:37

$\int \frac{\sqrt{\ln^3 x}}{x} \, {\rm d}x = \begin{vmatrix} \ln x = t \nl \frac{1}{x} {\rm d}x = {\rm d}t\end{vmatrix} = \int \sqrt{t^3}\, {\rm d}t = \int t^{\frac{3}{2}}\,{\rm d}t = \frac{2}{5}t^2 \sqrt{t} + C = \frac{2}{5} \ln^2 x \sqrt{\ln x} + C$

plisna · 08. 01. 2008 15:50 — Editoval plisna (08. 01. 2008 15:55)

$\int \frac{x-3}{x^2 -2x -5} \, {\rm d}x = \int \frac{x}{x^2-2x-5} \, {\rm d}x - \int \frac{3}{x^2-2x-5} \, {\rm d}x = \frac{1}{2} \int \frac{2x - 2 + 2}{x^2-2x-5} \, {\rm d}x - 3 \int \frac{1}{x^2-2x-5} \, {\rm d}x = \frac{1}{2} \int \frac{2x-2}{x^2-2x-5} \, {\rm d}x + \int \frac{1}{x^2-2x-5} \, {\rm d}x - \nl -3 \int \frac{1}{x^2-2x-5} \, {\rm d}x = \frac{1}{2} \ln |x^2-2x-5| - 2 \int \frac{1}{x^2-2x-5} \, {\rm d}x + C = \frac{1}{2} \ln |x^2-2x-5| -2 \int \frac{1}{(x-1)^2-6} \, {\rm d}x +C = \begin{vmatrix} x-1=t\nl{\rm d} x = {\rm d}t \end{vmatrix} = \frac{1}{2} \ln |x^2-2x-5| - \nl -2 \int \frac{1}{t^2-6} \, {\rm d}t +C= \begin{vmatrix} \sqrt{6}z = t \nl \sqrt{6} {\rm d}z = {\rm d}t \end{vmatrix} = \frac{1}{2} \ln |x^2-2x-5| - 2 \int \frac{\sqrt{6}}{6z^2 - 6} \, {\rm d}z +C= \frac{1}{2} \ln |x^2-2x-5| - \frac{\sqrt{6}}{3} \int \frac{1}{z^2-1} \, {\rm d}z=\frac{1}{2} \ln |x^2-2x-5| + \frac{\sqrt{6}}{6} \ln \left| \frac{|\frac{x-1}{\sqrt{6}}+1|}{|\frac{x-1}{\sqrt{6}}-1|} \right| + C$

Matematické Fórum

#1 08. 01. 2008 14:57

integral

#2 08. 01. 2008 15:37

Re: integral

#3 08. 01. 2008 15:50 — Editoval plisna (08. 01. 2008 15:55)

Re: integral

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