Was ist das Integral von # x / (1 + x ^ 2) #?

Lassen #u(x)=1+x^2" "# dann #" "du(x) =2xdx #
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#color(blue)((d(u(x)))/2=xdx)#
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Lösen Sie das Integral.
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#intx/(x^2+1)dx#
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#=intcolor(blue)((d(u(x)))/(2u(x))#
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#=1/2int(du(x))/(u(x))#
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#=1/2lnabs(u(x))+C#
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#=1/2lnabs(x^2+1)+C#
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weil #x^2+1>0 " " then " " abs(x^2+1)=x^2+1#
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Deswegen,
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#intx/(x^2+1)dx=1/2ln(x^2+1)+C#