Całka niewymierna

[bombel87]

Jak rozwiacac calke tego typu??Bo nie moge znalezc jakiegos przykladu ani wzoru ogolnego
\(\displaystyle{ \int \frac{dx}{(3-2x)\sqrt{x^2-4x+3}}}\)

[qaz]

spróbuj podstawiając: \(\displaystyle{ \frac{1}{3-2x}=t}\) z tego co pamiętam, to powinno wyjść ...

[Lider_M]

\(\displaystyle{ \int\frac{dx}{[1-2(x-2)]\sqrt{(x-2)^2-1}}=\int\frac{dt}{(1-2t)\sqrt{t^2-1}}}\)
I teraz \(\displaystyle{ \int\frac{dt}{(1-2t)(t+1)\sqrt{\frac{t-1}{t+1}}}}\) no i podstawienie \(\displaystyle{ \frac{t-1}{t+1}=m^2}\)

[jasny]

Założenia:
\(\displaystyle{ 3-2x\neq0\,\Leftrightarrow\,x\neq\frac{3}{2}}\)
\(\displaystyle{ x^2-4x+3>0}\)
\(\displaystyle{ (x-3)(x-1)>0}\)
\(\displaystyle{ x\in(-\infty;1)\cup(3;\infty)}\)

\(\displaystyle{ \frac{1}{3-2x}=t}\)
\(\displaystyle{ 3-2x=\frac{1}{t}}\)
\(\displaystyle{ x=\frac{3t-1}{2t}}\)
\(\displaystyle{ dx=\frac{6t-6t+2}{4t^2}dt=\frac{dt}{2t^2}}\)

\(\displaystyle{ I=\int\frac{tdt}{2t^2\sqrt{\frac{9t^2-6t+1}{4t^2}-\frac{12t-4}{2t}+3}}=\int\frac{dt}{2t\sqrt{\frac{9t^2-6t+1-24t^2+8t+12t^2}{4t^2}}}=\int\frac{dt}{\frac{t}{\sqrt{t^2}}\sqrt{-3t^2+2t+1}}}\)

Dla \(\displaystyle{ x\in(-\infty;1),\;t>0}\), więc \(\displaystyle{ I=\int\frac{dt}{\sqrt{-3t^2+2t+1}}}\)
Dla \(\displaystyle{ x\in(3;\infty),\;t}\)

[bombel87]

Thx