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Integral of (1/(x^3+x))dx?
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partial fractions! :)
\[\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+x(Bx+C)}{x(x^2+1)}\] => \[1=Ax^2+Bx^2+Cx+A\]
\[=>1=A, A+B=0,C=0=>B=-1\]
\[ \int 1/(x^3+x) dx\] \[ 1/(x^2+1)= 1/x- x/(x^2+1)\]
yep :)
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integrate each term \[\int 1/x = ln x\] \[ \int x/(x^2+1)= 1/2 * ln( x^2+1)\]
Thanks.
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