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

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

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Answer 1 · 2016-05-22T17:22:22

By converting the fraction into a sum of independent factors

Explanation:

We know how to integrate functions of the form $1/x^n$ or $1/(x-a)^n$ so we want to transform the complicated fraction into a sum of simple fractions. We write
$I = A/(x-1) + B/(x-1)^2 + C/(x^2 + 1)$
The constant term in the numerator is $B + C-A$ and is equal to $-1$
The $x^4$ term is $B+C+A$ and must be zero. This gives us
$-2A = -1$ or $A=-1/2$
The $x^3$ term is $-2C + A + B = 0$

Next $-2C + A +B +2C = 0 + 2(-A-B)$ so
$B = -A = 1/2$ and thus $C=0$

Thus $I = -(1/2)1/(x-1) + (1/2)1/(x-1)^2$

Integral $= -1/2 ln(x-1) -1/2(x-1)^2$

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

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Explanation:

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

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Explanation:

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