How do you integrate $int (x^2-1)/((x)*(x^2+1))$ using partial fractions?

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How do you integrate $int (x^2-1)/((x)*(x^2+1))$ using partial fractions?

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Answer 1 · 2016-10-18T13:08:02

$=ln((x^2+1)/x)+C$

Explanation:

$(x^2-1)/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)$
Developing
$(x^2-1)/(x(x^2+1))=(A(x^2+1)+ x(Bx+C))/(x(x^2+1)$
So we can compare the coefficients
$x^2-1=A(x^2+1)+ x(Bx+C)$
For $x^2$ , $1=A+B$ and $-1=A$ and $0=C$
Hence $A=-1$ and $B=2$
$int((x^2-1)dx)/(x(x^2+1))=intAdx/x+int((Bx+C)dx)/(x^2+1)$
$=int(-1dx)/x+int((2x)dx)/(x^2+1$
$=-lnx +ln(x^2+1)+C$
$=ln((x^2+1)/x)+C$

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How do you integrate $int (x^2-1)/((x)*(x^2+1))$ using partial fractions?

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How do you integrate $int (x^2-1)/((x)*(x^2+1))$ using partial fractions?