Całki
: 27 sie 2007, o 09:41
Proszę o pomoc z tymi całkami:
1) \(\displaystyle{ \int {\frac{{dx}}{{\sqrt[3]{{3x + 1}} - 1}}}}\)
2) \(\displaystyle{ \int {\frac{{dx}}{{\sqrt {2x^2 + 4x + 3} }}}}\)
3) \(\displaystyle{ \int {\frac{{x^2 }}{{2x^3 + 2x^2 + 2x}}}}\)
A w tej całce nie wiem jak to skończyć. I nie wiem czy dobrze licze ją
4) \(\displaystyle{ \int {\frac{{x^5 - 1}}{{x^4 - 5x^2 + 6}}} dx}\)
\(\displaystyle{ \int {\frac{{x^5 - 1}}{{x^4 - 5x^2 + 6}}} dx = t {xdx} + t {\frac{{5x^3 - 6x - 1}}{{x^4 - 5x^2 + 6}}} dx =}\)
\(\displaystyle{ \int {xdx} = \frac{{x^2 }}{2}}\)
\(\displaystyle{ \int {\frac{{5x^3 - 6x - 1}}{{x^4 - 5x^2 + 6}}} dx = |x^2 = t \to ft( {x^2 - 2} \right)\left( {x^2 - 3} \right)}\)
\(\displaystyle{ \int {\frac{{5x^3 - 6x - 1}}{{x^4 - 5x^2 + 6}}} dx = t {\frac{{Ax + B}}{{x^2 - 2}}} + t {\frac{{Cx + D}}{{x^2 - 3}}}}\)
\(\displaystyle{ 5x^3 - 6x - 1 = ft( {Ax + B} \right)\left( {x^2 - 3} \right) + ft( {Cx + D} \right)\left( {x^2 - 2} \right)}\)
gdy \(\displaystyle{ x = - \sqrt 2 i}\)
\(\displaystyle{ A = - 16}\)
\(\displaystyle{ B = 0}\)
...
\(\displaystyle{ - 16\int {\frac{{dx}}{{x^2 - 2}}} - t {\frac{{21x + 2}}{{x^2 - 3}}}}\)
Jak to dalej rozpisać ?
1) \(\displaystyle{ \int {\frac{{dx}}{{\sqrt[3]{{3x + 1}} - 1}}}}\)
2) \(\displaystyle{ \int {\frac{{dx}}{{\sqrt {2x^2 + 4x + 3} }}}}\)
3) \(\displaystyle{ \int {\frac{{x^2 }}{{2x^3 + 2x^2 + 2x}}}}\)
A w tej całce nie wiem jak to skończyć. I nie wiem czy dobrze licze ją
4) \(\displaystyle{ \int {\frac{{x^5 - 1}}{{x^4 - 5x^2 + 6}}} dx}\)
\(\displaystyle{ \int {\frac{{x^5 - 1}}{{x^4 - 5x^2 + 6}}} dx = t {xdx} + t {\frac{{5x^3 - 6x - 1}}{{x^4 - 5x^2 + 6}}} dx =}\)
\(\displaystyle{ \int {xdx} = \frac{{x^2 }}{2}}\)
\(\displaystyle{ \int {\frac{{5x^3 - 6x - 1}}{{x^4 - 5x^2 + 6}}} dx = |x^2 = t \to ft( {x^2 - 2} \right)\left( {x^2 - 3} \right)}\)
\(\displaystyle{ \int {\frac{{5x^3 - 6x - 1}}{{x^4 - 5x^2 + 6}}} dx = t {\frac{{Ax + B}}{{x^2 - 2}}} + t {\frac{{Cx + D}}{{x^2 - 3}}}}\)
\(\displaystyle{ 5x^3 - 6x - 1 = ft( {Ax + B} \right)\left( {x^2 - 3} \right) + ft( {Cx + D} \right)\left( {x^2 - 2} \right)}\)
gdy \(\displaystyle{ x = - \sqrt 2 i}\)
\(\displaystyle{ A = - 16}\)
\(\displaystyle{ B = 0}\)
...
\(\displaystyle{ - 16\int {\frac{{dx}}{{x^2 - 2}}} - t {\frac{{21x + 2}}{{x^2 - 3}}}}\)
Jak to dalej rozpisać ?