How do you evaluate the integral #int (x+5)/(3x-1)#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Narad T. Feb 20, 2017 The answer is #=1/3x+16/9ln(|3x+1|)+C# Explanation: We need #intdx/x=lnx+C# Let's rewrite #x+5=1/3(3x-1)+16/3# So, #((x+5))/(3x+1)=1/3*cancel(3x-1)/cancel(3x-1)+16/3*1/(3x+1)# #int((x+5)dx)/(3x+1)=int1/3dx+16/3intdx/(3x+1)# #=1/3x+16/3ln(|3x+1|)/3+C# #=1/3x+16/9ln(|3x+1|)+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 1315 views around the world You can reuse this answer Creative Commons License