How do you integrate #int ( x-2 ) / (x^2 + 4x + 3)# using partial fractions?
1 Answer
Feb 12, 2016
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#
Explanation:
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#