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

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

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Answer 1 · 2016-10-18T13:08:02

$= ln (\frac{x^{2} + 1}{x}) + C$ $=ln((x^2+1)/x)+C$

Explanation:

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

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

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