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`