How do you integrate $int (x^2)/(x^2 + 3x -4) dx$ using partial fractions?

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Answer 1 · 2018-05-13T21:58:19

$int x^2/(x^2+3x-4)dx = x+1/5lnabs(x+1) -16/5lnabs(x+4)+C$

Before applying partial fraction decomposition we must reduce the degree of the numerator:

$x^2/(x^2+3x-4) = (x^2+3x-4 -3x + 4)/(x^2+3x-4) = 1- (3x-4)/(x^2+3x-4)$

Factorize the denominator:

$x^2+3x-4 = (x-1)(x+4)$

now decompose in partial fractions:

$(3x-4)/(x^2+3x-4) = A/(x-1)+B/(x+4)$

$(3x-4)/(x^2+3x-4) = (A(x+4) +B(x-1) )/(x^2+3x-4)$

$(3x-4) = (A+B)x + (4A -B)$

${(A+B=3),(4A-B= -4):}$

${(A=-1/5),(B= 16/5):}$

$(3x-4)/(x^2+3x-4) = -1/(5(x-1))+16/(5(x+4))$

Now:

$int x^2/(x^2+3x-4)dx = int (1+1/5 1/(x-1) -16/5 1/(x+4))dx$

and using the linearity of the integral:

$int x^2/(x^2+3x-4)dx = int dx+1/5 int dx/(x-1) -16/5 int dx/(x+4)$

$int x^2/(x^2+3x-4)dx = x+1/5lnabs(x+1) -16/5lnabs(x+4)+C$

How do you integrate int (x^2)/(x^2 + 3x -4) dxint (x^2)/(x^2 + 3x -4) dx using partial fractions?