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

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Answer 1 · 2016-02-12T07:21:27

Decompose the integrand into partial fractions to find:

#int (x-2)/(x^2+4x+3) dx =-3/2 ln(abs(x+1)) + 5/2 ln(abs(x+3)) + C#

Let's find the partial fraction decomposition first:

#(x-2)/(x^2+4x+3)#

#=(x-2)/((x+1)(x+3))#

#=A/(x+1)+B/(x+3)#

#=(A(x+3)+B(x+1))/((x+1)(x+3))#

#=((A+B)x+(3A+B))/((x+1)(x+3))#

So:

#(A+B)x+(3A+B) = x-2#

Hence:

#A+B = 1#

#3A+B = -2#

So:

#A = -3/2#

#B = 5/2#

Putting it together:

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

Then use:

#int(1/t) dt = ln abs(t) + C#

to find:

#int (-3/(2(x+1)) + 5/(2(x+3))) dx#

#=-3/2 ln(abs(x+1)) + 5/2 ln(abs(x+3)) + C#