How do you integrate #(2x-82)/(x^2+2x-48) dx# using partial fractions? Calculus Techniques of Integration Integral by Partial Fractions 1 Answer Narad T. Oct 15, 2016 7ln(x+8)-5ln(x-6)+C Explanation: #int(2x-82)dx/(x^2+2x-48)# #(2x-82)/(x^2+2x-48)=(2x-82)/((x-6)(x+8))=A/(x-6)+B/(x+8) =(A(x+8) +B(x-6))/((x-6)(x+8))# #2x-82=A(x+8) +B(x-6)# #if x=6 ; -70=14A ; A=-5# #if x=-8 ; 98=-14B , B=7# #int(2x-82)dx/(x^2+2x-48)=int7dx/(x+8)-int5dx/(x-6# #=7ln(x+8)-5ln(x-6)+C# Answer link Related questions How do I find the partial fraction decomposition of #(2x)/((x+3)(3x+1))# ? How do I find the partial fraction decomposition of #(1)/(x^3+2x^2+x# ? How do I find the partial fraction decomposition of #(x^4+1)/(x^5+4x^3)# ? How do I find the partial fraction decomposition of #(x^4)/(x^4-1)# ? How do I find the partial fraction decomposition of #(t^4+t^2+1)/((t^2+1)(t^2+4)^2)# ? How do I find the integral #intt^2/(t+4)dt# ? How do I find the integral #int(x-9)/((x+5)(x-2))dx# ? How do I find the integral #int1/((w-4)(w+1))dw# ? How do I find the integral #intdx/(x^2(x-1)^2)# ? How do I find the integral #int(x^3+4)/(x^2+4)dx# ? See all questions in Integral by Partial Fractions Impact of this question 1268 views around the world You can reuse this answer Creative Commons License