How do you integrate #int4/(x^2 + 9)dx#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Sasha P. Sep 13, 2015 #4/3arctan(x/3)+C# Explanation: Let #x=3t#, then: #dx=3dt,# #int4/(x^2+9)dx=int4/(9t^2+9)3dt=int4/9(t^2+1)3dt=# #=4/3intdt/(t^2+1)=4/3arctant+C=4/3arctan(x/3)+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 2172 views around the world You can reuse this answer Creative Commons License