Begin by factoring the denominator to simplify the integral to:
int(5x+11)/((x+7)(x-5))dx∫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)Ax+7+Bx−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)=Ax+7+Bx−5
(5x+11)/((x+7)(x-5))=(A(x-5)+B(x+7))/((x+7)(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)5x+11=A(x−5)+B(x+7)
Set x=5x=5 to find the value of BB:
5(5)+11=A(5-5)+B(5+7)->36=12B->B=35(5)+11=A(5−5)+B(5+7)→36=12B→B=3
Similarly, set x=-7x=−7 to find AA:
5(-7)+11=A(-7-5)+B(-7+7)->-24=-12A->A=25(−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)5x+11(x+7)(x−5)=2x+7+3x−5
Putting this back into the integral and evaluating:
int(5x+11)/((x+7)(x-5))dx=int2/(x+7)+3/(x-5)dx∫5x+11(x+7)(x−5)dx=∫2x+7+3x−5dx
color(white)(XX)=int2/(x+7)dx+int3/(x-5)dxXX=∫2x+7dx+∫3x−5dx
color(white)(XX)=2int1/(x+7)dx+3int1/(x-5)dxXX=2∫1x+7dx+3∫1x−5dx
color(white)(XX)=2lnabs(x+7)+3lnabs(x-5)+CXX=2ln|x+7|+3ln|x−5|+C
