How do you integrate #int (x^2 +x -16) / [(x+1)(x-3)^2] dx# using partial fractions?

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Answer 1 · 2017-08-04T21:29:21

# int \ (x^2+x-16)/((x+1)(x-3)^2) \ dx = -ln|x+1| + 2 ln|x-3| + 1/(x-3) + C #

We have:

# I = int \ (x^2+x-16)/((x+1)(x-3)^2) \ dx #

We can decompose the integrand into partial fraction, which will take the form:

# (x^2+x-16)/((x+1)(x-3)^2) -= A/(x+1)+B/(x-3) + C/(x-3)^2 #
# " " = (A(x-3)^2+B(x+1)(x-3) + C(x+1))/((x+1)(x-3)^2) #

Leading to the identity:

# x^2+x-16 -= A(x-3)^2+B(x+1)(x-3) + C(x+1) #

Where #A,B,C# are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put #x=-1 => -16=A(-4)^2 => A=-1 #
Put #x= \ \ \ \ 3 => -4=4C=> C=-1#
Put #x= \ \ \ \ 0 => -16=9A-3B+C => B=2#

So using partial fraction decomposition we have:

# I = int \ -1/(x+1)+2/(x-3) - 1/(x-3)^2 \ dx #
# \ \ = int \ -1/(x+1) \ dx +2 \ int \ 1/(x-3) \ dx - int \ 1/(x-3)^2 \ dx #

And now all integrands are readily integratable, so:

# I = -ln|x+1| + 2 ln|x-3| + 1/(x-3) + C #