How to do integrate $int(5x^3-x^2)/(x^2-1)dx$ ?

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how do you solve by using partial fractions?

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Answer 1 · 2018-04-17T23:42:08

$int (5x^3-x^2)/(x^2-1)dx = 5/2 x^2-x+3ln abs(x+1)+2ln abs(x-1)+C$

Reduce the degree of the numerator:

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

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

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

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

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

Use now the partial fraction decomposition to the resulting proper rational function. Factorize the denominator:

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

$(5x-1)/(x^2-1) = A/(x+1)+B/(x-1)$

$(5x-1)/(x^2-1) = (A(x-1)+B(x+1))/(x^2-1)$

$5x-1 = (A+B)x - (A-B)$

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

${(A= 3),(B=2):}$

Then:

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

Using the linearity of the integral then:

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

$int (5x^3-x^2)/(x^2-1)dx = 5/2 x^2-x+3ln abs(x+1)+2ln abs(x-1)+C$