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