Witam, mam problem z tymi dwiema całkami:
1) \(\int \frac{ \sqrt{1-4x-x^2} }{x-1} dx\)
2)\(\int \frac{ln^2x}{x^2} dx\)
całki nieoznaczone
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Crazy Driver
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Re: całki nieoznaczone
2) \(x=e^t\)
\(dx=e^t\,dt\)
\(\int \frac{\ln^2x}{x}\,dx=\int \frac{t^2}{e^{2t}}\cdot e^t\,dt=\int \frac{t^2}{e^t}\,dt=\int t^2e^{-t}\,dt\)
\(v=-t\)
\(dt=-dv\)
\(\int t^2e^{-t}\,dt=-\int v^2e^v\,dv=-\int v^2(e^v)'\,dv=-\left(v^2e^v-\int 2ve^v\,dv\right)=-v^2e^v+2\int ve^v\,dv=\\\\=-v^2e^v+2\left(ve^v-\int e^v\,dv\right)=-v^2e^v+2ve^v-2e^v+C=-t^2e^{-t}-2te^{-t}-2e^{-t}+C=\\\\=- \frac{\ln^2x}{x^2}- \frac{2\ln x}{x}- \frac{2}{x}+C\)
\(dx=e^t\,dt\)
\(\int \frac{\ln^2x}{x}\,dx=\int \frac{t^2}{e^{2t}}\cdot e^t\,dt=\int \frac{t^2}{e^t}\,dt=\int t^2e^{-t}\,dt\)
\(v=-t\)
\(dt=-dv\)
\(\int t^2e^{-t}\,dt=-\int v^2e^v\,dv=-\int v^2(e^v)'\,dv=-\left(v^2e^v-\int 2ve^v\,dv\right)=-v^2e^v+2\int ve^v\,dv=\\\\=-v^2e^v+2\left(ve^v-\int e^v\,dv\right)=-v^2e^v+2ve^v-2e^v+C=-t^2e^{-t}-2te^{-t}-2e^{-t}+C=\\\\=- \frac{\ln^2x}{x^2}- \frac{2\ln x}{x}- \frac{2}{x}+C\)
Ostatnio zmieniony 16 sty 2012, 04:35 przez Crazy Driver, łącznie zmieniany 1 raz.
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Crazy Driver
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- Rejestracja: 07 maja 2010, 12:48
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Re: całki nieoznaczone
\(\int \frac{\sqrt{-x^2-4x+1}}{x-1}\,dx=\int \frac{-x^2-4x+1}{(x-1)\sqrt{-x^2-4x+1}}\,dx=\int \frac{(x-1)(-x-5)-4}{(x-1)\sqrt{-x^2-4x+1}}\,dx=\\\\=-\underbrace{\int \frac{x+5}{\sqrt{-x^2-4x+1}}\,dx}_{I}-4\underbrace{\int \frac{dx}{(x-1)\sqrt{-x^2-4x+1}}}_{J}\)
\(I=\int \frac{x+5}{\sqrt{-x^2-4x+1}}\,dx=-\int \frac{-2x-4}{2\sqrt{-x^2-4x+1}}\,dx+3\underbrace{\int \frac{dx}{\sqrt{-x^2-4x+1}}}_{K}=-\int \frac{(-x^2-4x+1)'}{2\sqrt{-x^2-4x+1}}\,dx+3K=\\\\=-\sqrt{-x^2-4x+1}+3K\)
\(K=\int \frac{dx}{\sqrt{-x^2-4x+1}}=\int \frac{dx}{\sqrt{-(x+2)^2+5}\)
\(t=x+2\)
\(dt=dx\)
\(\int \frac{dx}{\sqrt{-(x+2)^2+5}}=\int \frac{dt}{\sqrt{5-t^2}}\)
\(t=u\sqrt5\)
\(dt=\sqrt5\,du\)
\(\int \frac{dt}{\sqrt{5-t^2}}=\int \frac{\sqrt5\,du}{\sqrt{5-5u^2}}=\int \frac{du}{\sqrt{1-u^2}}=\arcsin u+C_1=\arcsin \frac{t}{\sqrt5}+C_1=\arcsin \frac{x+2}{\sqrt5}+C_1=K\)
\(J=\int \frac{dx}{(x-1)\sqrt{-x^2-4x+1}}\)
\(v= \frac{1}{x-1}\)
\(x-1= \frac{1}{v}\)
\(dv= -\frac{dx}{(x-1)^2}\)
\(\int \frac{dx}{(x-1)\sqrt{-x^2-4x+1}}=\int \frac{x-1}{(x-1)^2\sqrt{-(x+2)^2+5}}\,dx=-\int\frac{dv}{ v \sqrt{-\left( \frac{1}{v}+3\right)^2+5 }}=- \int\frac{dv}{\sqrt{-4v^2-6v-1}}=\\\\=-\int \frac{dv}{\sqrt{-4\left(v+ \frac{3}{4}\right)^2+ \frac{5}{4}} }\)
\(p=v+ \frac{3}{4}\)
\(dp=dv\)
\(-\int \frac{dv}{\sqrt{-4\left(v+ \frac{3}{4}\right)^2+ \frac{5}{4}} }=-\int \frac{dp}{\sqrt{-4p^2+ \frac{5}{4}} }\)
\(w=\frac{4}{\sqrt{5}}p\)
\(dp= \frac{\sqrt5}{4}\,dw\)
\(-\int \frac{dp}{\sqrt{-4p^2+ \frac{5}{14}} }= -\frac{\sqrt5}{4}\int \frac{dw}{\sqrt{- \frac{5}{4} w^2+ \frac{5}{4}} }= -\frac{1}{2}\int \frac{dw}{\sqrt{1-w^2}}= -\frac{1}{2}\arcsin w+C_2=-\frac{1}{2}\arcsin \frac{4v+3}{\sqrt5} +C_2=\)
\(=-\frac{1}{2}\arcsin \frac{\sqrt5(3x+1)}{x-1} +C_2=J\)
\(\int \frac{\sqrt{-x^2-4x+1}}{x-1}\,dx=-I-4J=-(-\sqrt{-x^2-4x+1}+3K)-4J=\sqrt{-x^2-4x+1}-3K-4J=\)
\(=\sqrt{-x^2-4x+1}-3\arcsin \frac{x+2}{\sqrt5}+2\arcsin \frac{\sqrt5(3x+1)}{x-1}+C\)
\(I=\int \frac{x+5}{\sqrt{-x^2-4x+1}}\,dx=-\int \frac{-2x-4}{2\sqrt{-x^2-4x+1}}\,dx+3\underbrace{\int \frac{dx}{\sqrt{-x^2-4x+1}}}_{K}=-\int \frac{(-x^2-4x+1)'}{2\sqrt{-x^2-4x+1}}\,dx+3K=\\\\=-\sqrt{-x^2-4x+1}+3K\)
\(K=\int \frac{dx}{\sqrt{-x^2-4x+1}}=\int \frac{dx}{\sqrt{-(x+2)^2+5}\)
\(t=x+2\)
\(dt=dx\)
\(\int \frac{dx}{\sqrt{-(x+2)^2+5}}=\int \frac{dt}{\sqrt{5-t^2}}\)
\(t=u\sqrt5\)
\(dt=\sqrt5\,du\)
\(\int \frac{dt}{\sqrt{5-t^2}}=\int \frac{\sqrt5\,du}{\sqrt{5-5u^2}}=\int \frac{du}{\sqrt{1-u^2}}=\arcsin u+C_1=\arcsin \frac{t}{\sqrt5}+C_1=\arcsin \frac{x+2}{\sqrt5}+C_1=K\)
\(J=\int \frac{dx}{(x-1)\sqrt{-x^2-4x+1}}\)
\(v= \frac{1}{x-1}\)
\(x-1= \frac{1}{v}\)
\(dv= -\frac{dx}{(x-1)^2}\)
\(\int \frac{dx}{(x-1)\sqrt{-x^2-4x+1}}=\int \frac{x-1}{(x-1)^2\sqrt{-(x+2)^2+5}}\,dx=-\int\frac{dv}{ v \sqrt{-\left( \frac{1}{v}+3\right)^2+5 }}=- \int\frac{dv}{\sqrt{-4v^2-6v-1}}=\\\\=-\int \frac{dv}{\sqrt{-4\left(v+ \frac{3}{4}\right)^2+ \frac{5}{4}} }\)
\(p=v+ \frac{3}{4}\)
\(dp=dv\)
\(-\int \frac{dv}{\sqrt{-4\left(v+ \frac{3}{4}\right)^2+ \frac{5}{4}} }=-\int \frac{dp}{\sqrt{-4p^2+ \frac{5}{4}} }\)
\(w=\frac{4}{\sqrt{5}}p\)
\(dp= \frac{\sqrt5}{4}\,dw\)
\(-\int \frac{dp}{\sqrt{-4p^2+ \frac{5}{14}} }= -\frac{\sqrt5}{4}\int \frac{dw}{\sqrt{- \frac{5}{4} w^2+ \frac{5}{4}} }= -\frac{1}{2}\int \frac{dw}{\sqrt{1-w^2}}= -\frac{1}{2}\arcsin w+C_2=-\frac{1}{2}\arcsin \frac{4v+3}{\sqrt5} +C_2=\)
\(=-\frac{1}{2}\arcsin \frac{\sqrt5(3x+1)}{x-1} +C_2=J\)
\(\int \frac{\sqrt{-x^2-4x+1}}{x-1}\,dx=-I-4J=-(-\sqrt{-x^2-4x+1}+3K)-4J=\sqrt{-x^2-4x+1}-3K-4J=\)
\(=\sqrt{-x^2-4x+1}-3\arcsin \frac{x+2}{\sqrt5}+2\arcsin \frac{\sqrt5(3x+1)}{x-1}+C\)
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