How do you use partial fraction decomposition to decompose the fraction to integrate #(-x - 38)/(2x^2 + 9x - 5) #?

1 Answer
Nov 29, 2016

# int(-x-38)/(2x^2+9x-5) dx= 3ln|x+5| -7/2ln|2x-1| + c#

Explanation:

First we need to factorise the denominator.

To do this we need to find factors of #-10# (coefficient of #x^2 xx# coefficient of constant#= 2 xx -5#) that add up tp 9 (the coefficient of #x=9#). So the factors we seek are #-1# and #10#

# :. 2x^2+9x-5 = 2x^2 +10x - x -5 #
# :. 2x^2+9x-5 =2x(x+5) - (x+5) #
# :. 2x^2+9x-5 =(x+5)(2x-1) #

Therefore we can write the integrand as follows:

# (-x-38)/(2x^2+9x-5) = (-x-38)/((x+5)(2x-1))#

And thge partial fraction decomposition will be:

# \ \ \ \ \ (-x-38)/(2x^2+9x-5) = A/(x+5) + B/(2x-1)#
# :. (-x-38)/(2x^2+9x-5) = (A(2x-1)+B(x+5))/((x+5)(2x-1)) #
# :. \ \ \ \ (-x-38) = A(2x-1)+B(x+5) #

Subs #x=-5 => -(-5)-38=A(-10-1) + 0#
# :. -11A=-33 => A=3 #

And Sub #x=1/2=>-1/2-38 = B(1/2+5) #
# :. 11/2B=-77/2 => B=-7#

Hence, the partial fraction decomposition of the integrand is:

# (-x-38)/(2x^2+9x-5) = 3/(x+5) -7/(2x-1)#

And so;

# int(-x-38)/(2x^2+9x-5) dx= int 3/(x+5) -7/(2x-1) dx#

And integrating we get:

# int(-x-38)/(2x^2+9x-5) dx= 3ln|x+5| -7/2ln|2x-1| + c#