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Obliczenie całki
: 15 maja 2019, o 20:36
autor: aneta909811
Jak obliczyć takiego byczka \(\displaystyle{ \int x \frac{ \sqrt{x-2} }{ \sqrt{4-x} } \mbox{d}x}\)
Próbowałam przez części, ale jakoś nie wychodzi...
Re: Obliczenie całki
: 15 maja 2019, o 20:38
autor: Mariusz M
Spróbuj podstawienia
\(\displaystyle{ t = \frac{ \sqrt{x-2} }{ \sqrt{4-x} }}\)
chyba że musisz przez części
\(\displaystyle{ \int x \frac{ \sqrt{x-2} }{ \sqrt{4-x} } \mbox{d}x\\
t=\frac{ \sqrt{x-2} }{ \sqrt{4-x} }\\
t^2=\frac{x-2}{4-x}\\
t^2=\frac{x-4+2}{4-x}\\
t^2=-1+\frac{2}{4-x}\\
t^2+1=\frac{2}{4-x}\\
\frac{t^2+1}{2}= \frac{1}{4-x}\\
\frac{2}{t^2+1} =4-x\\
x-4=-\frac{2}{t^2+1}\\
x = 4 -\frac{2}{t^2+1}\\
x=\frac{4t^2+4-2}{t^2+1}\\
x= \frac{4t^2+2}{t^2+1} \\
\mbox{d}x =\left( -2\right)\left( -1\right) \left( t^2+1\right)^{-2} \cdot 2t \mbox{d}t\\
\mbox{d}x = \frac{4t}{\left( t^2+1\right)^2 } \mbox{d}t\\
\int{ \frac{4t^2+2}{t^2+1} \cdot t \cdot \frac{4t}{\left( t^2+1\right)^2 } \mbox{d}t} \\
8 \int{ \frac{t^2\left( 2t^2+1\right) }{\left( t^2+1\right)^3 } \mbox{d}t } \\}\)
i teraz albo wydzielenie części wymiernej sposobem Ostrogradskiego
albo zastosowanie wzoru redukcyjnego
\(\displaystyle{ 8\int{\frac{2t^4+t^2}{\left( t^2+1\right)^3 }\mbox{d}t}=8\int{\frac{\left(2t^4+4t^2+2 \right)-3\left( t^2+1\right)+1 }{\left( t^2+1\right)^3 } \mbox{d}t}\\
=16\int{ \frac{ \mbox{d}t }{t^2+1} }-24 \int{\frac{ \mbox{d}t}{\left( t^2+1\right)^2 }} +8 \int{ \frac{ \mbox{d}t}{\left( t^2+1\right)^3 } }}\)
Wzór redukcyjny
\(\displaystyle{ \int{\frac{ \mbox{d}x }{\left( x^2+1\right)^n }}=\int{\frac{1+x^2-x^2}{\left( x^2+1\right)^n } \mbox{d}x }\\
=\int{ \frac{1+x^2}{\left( x^2+1\right)^n } \mbox{d}x }-\int{ \frac{x^2}{\left( x^2+1\right)^n } \mbox{d}x }\\
=\int{ \frac{ \mbox{d}x }{\left( x^2+1\right)^{n-1} } }-\int{ \frac{x^2}{\left( x^2+1\right)^n } \mbox{d}x }\\
=\int{ \frac{ \mbox{d}x }{\left( x^2+1\right)^{n-1} } }+\int{ \frac{x}{\left( 2n-2\right) } \cdot \frac{\left( -1\right)\left( 2n-2\right)x }{\left( x^2+1\right)^n } \mbox{d}x }\\
=\int{ \frac{ \mbox{d}x }{\left( x^2+1\right)^{n-1} } }+\frac{1}{2n-2} \frac{x}{\left( x^2+1\right)^{n-1} } -\frac{1}{2n-2}\int{ \frac{ \mbox{d}x }{\left( x^2+1\right)^{n-1} } }\\
=\frac{1}{2n-2} \frac{x}{\left( x^2+1\right)^{n-1} }+ \frac{2n-3}{2n-2}\int{ \frac{ \mbox{d}x }{\left( x^2+1\right)^{n-1} } }\\
\int{\frac{ \mbox{d}x }{\left( x^2+1\right)^n }}=\frac{1}{2n-2} \frac{x}{\left( x^2+1\right)^{n-1} }+ \frac{2n-3}{2n-2}\int{ \frac{ \mbox{d}x }{\left( x^2+1\right)^{n-1} } }\\}\)