How do you integrate #(x^2-2)/((x+1)(x^2+3))# using partial fractions?

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Answer 1 · 2016-10-29T21:47:38

Please see the explanation.

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#(x^2 - 2)/((x + 1)(x^2 + 3)) = A/(x + 1) + (Bx + C)/(x^2 + 3)#

#x^2 - 2 = A(x^2 + 3) + (Bx + C)(x + 1)#

Let x = -1 to make B and C disappear:

#-1^2 - 2 = A(-1^2 + 3)#

#-1 = 4A#

#A = -1/4#

#x^2 - 2 = -1/4(x^2 + 3) + (Bx + C)(x + 1)#

Let x = 0 to make B disappear:

#- 2 = -1/4(3) + (C)(1)#

#C = -5/4#

#x^2 - 2 = -1/4(x^2 + 3) + (Bx -5/4)(x + 1)#

Let x = 1:

#1^2 - 2 = -1/4(1^2 + 3) + (B -5/4)(1 + 1)#

#-1 = -1 + (2B -5/2)#

#B = 5/4#

#int(x^2 - 2)/((x + 1)(x^2 + 3))dx = -1/4int1/(x + 1)dx + 5/4intx/(x^2 + 3)dx - 5/4int1/(x^2 + 3)dx#

#int(x^2 - 2)/((x + 1)(x^2 + 3))dx = -1/4ln|x + 1| + 5/4ln|x^2 + 3| - (5sqrt(3))/12tan^-1(x/sqrt(3))#