How do you evaluate the integral #int (2x-3)/((x-1)(x+4))#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Andrea S. Mar 8, 2018 #int (2x-3)/((x-1)(x+4)) dx = -1/5 ln abs (x-1)+11/5 ln abs (x+4) +C# Explanation: Use partial fractions decomposition: #(2x-3)/((x-1)(x+4)) = A/(x-1)+B/(x+4)# #2x-3 = A(x+4)+B(x-1)# #2x-3 = (A+B)x +4A-B# #{(A+B=2),(4A-B=-3):}# #{(5A=-1),(B=-A+2):}# #{(A=-1/5),(B=11/5):}# #(2x-3)/((x-1)(x+4)) = -1/(5(x-1))+11/(5(x+4))# #int (2x-3)/((x-1)(x+4)) dx = -1/5 int dx/(x-1)+11/5 int dx/(x+4)# #int (2x-3)/((x-1)(x+4)) dx = -1/5 ln abs (x-1)+11/5 ln abs (x+4) +C# Answer link Related questions How do you find the integral #int1/(x^2*sqrt(x^2-9))dx# ? How do you find the integral #intx^3/(sqrt(x^2+9))dx# ? How do you find the integral #intx^3*sqrt(9-x^2)dx# ? How do you find the integral #intx^3/(sqrt(16-x^2))dx# ? How do you find the integral #intsqrt(x^2-1)/xdx# ? How do you find the integral #intsqrt(x^2-9)/x^3dx# ? How do you find the integral #intx/(sqrt(x^2+x+1))dx# ? How do you find the integral #intdt/(sqrt(t^2-6t+13))# ? How do you find the integral #intx*sqrt(1-x^4)dx# ? How do you prove the integral formula #intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C# ? See all questions in Integration by Trigonometric Substitution Impact of this question 1376 views around the world You can reuse this answer Creative Commons License