How do you find #int (5x+11)/(x^2+2x-35) dx# using partial fractions?

Question

How do you find #int  (5x+11)/(x^2+2x-35)  dx# using partial fractions?

Cameron Alexander · Accepted Answer

Do a partial fraction decomposition on #(5x+11)/(x^2+2x-35)# and simplify to get #2lnabs(x+7)+3lnabs(x-5)+C#.

Begin by factoring the denominator to simplify the integral to: #int(5x+11)/((x+7)(x-5))dx# Because the denominator contains only linear factors, our partial fraction decomposition will be of the form: #A/(x+7)+B/(x-5)# We can now set up the decomposition: #(5x+11)/((x+7)(x-5))=A/(x+7)+B/(x-5)# #(5x+11)/((x+7)(x-5))=(A(x-5)+B(x+7))/((x+7)(x-5))# Equating the numerators and simplifying: #5x+11=A(x-5)+B(x+7)# Set #x=5# to find the value of #B#: #5(5)+11=A(5-5)+B(5+7)->36=12B->B=3# Similarly, set #x=-7# to find #A#: #5(-7)+11=A(-7-5)+B(-7+7)->-24=-12A->A=2# Our decomposition is therefore: #(5x+11)/((x+7)(x-5))=2/(x+7)+3/(x-5)# Putting this back into the integral and evaluating: #int(5x+11)/((x+7)(x-5))dx=int2/(x+7)+3/(x-5)dx# #color(white)(XX)=int2/(x+7)dx+int3/(x-5)dx# #color(white)(XX)=2int1/(x+7)dx+3int1/(x-5)dx# #color(white)(XX)=2lnabs(x+7)+3lnabs(x-5)+C#

William Abdalla · Answer

undefined To find ∫(5x+11)/(x^2+2x-35) dx using partial fractions, first factor the denominator: x^2 + 2x - 35 = (x + 7)(x - 5). Then, express the given fraction as A/(x + 7) + B/(x - 5). Next, find the values of A and B by equating numerators and simplifying. Once you have A and B, integrate each term separately. The integral of A/(x + 7) with respect to x is A ln|x + 7| + C, and the integral of B/(x - 5) with respect to x is B ln|x - 5| + C. Therefore, the integral of (5x+11)/(x^2+2x-35) dx using partial fractions is A ln|x + 7| + B ln|x - 5| + C.