How do you integrate $int 2/(x^3-x^2)$ using partial fractions?

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Answer 1 · 2016-12-16T22:08:55

$=2(ln|x - 1| - ln|x| + 1/x) + C$

$x^3 - x^2$ can be factored as $x^2(x - 1)$ .

$(Ax + B)/(x^2) + C/(x - 1) = 2/((x^2)(x - 1))$

$(Ax + B)(x- 1) + C(x^2) = 2$

$Ax^2 + Bx - Ax - B + Cx^2 = 2$

$(A + C)x^2 + (B - A)x + (-B) = 2$

We can now write a system of equations.

${(A + C = 0), (B - A = 0), (-B = 2):}$

Solving, we obtain $B = -2$ , $A = -2$ , $C = 2$ .

Therefore, the partial fraction decomposition is:

$(-2x - 2)/x^2 + 2/(x - 1)$

We now decompose $(-2x - 2)/x^2$ into partial fractions.

$A/x^2 + B/x = (-2x- 2)/x^2$

$A + Bx = -2x - 2 -> A = -2 and B = -2$

Therefore, the complete partial fraction decomposition of $2/(x^3 - x^2)$ is $-2/x^2 - 2/x + 2/(x - 1)$ .

This can be integrated as follows:

$int(-2/x^2 - 2/x + 2/(x - 1))dx$

$=-2ln|x| + 2ln|x- 1| - int(2x^-2) + C$

$=2ln|x - 1| - 2ln|x| -(-2x^-1) + C$

$=2ln|x- 1| - 2ln|x| + 2/x + C$

$=2(ln|x - 1| - ln|x| + 1/x) + C$

Hopefully this helps!

How do you integrate int 2/(x^3-x^2)int 2/(x^3-x^2) using partial fractions?