How do you integrate #int dx/(4x^2+9)^2# using trig substitutions? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Cem Sentin Mar 21, 2018 #int dx/(4x^2+9)^2=1/108arctan((2x)/3)+x/(72x^2+162)+C# Explanation: #int dx/(4x^2+9)^2# =#1/2int (2dx)/((2x)^2+3^2)^2# After using #2x=3tanu# and #2dx=3(secu)^2*du# transforms, this integral became #1/2int (3(secu)^2*du)/(81(secu)^4)# =#1/54int (cosu)^2*du# =#1/108int (1+cos2u)*du# =#1/108u+1/216sin2u+C# =#1/108u+1/216*(2tanu)/((tanu)^2+1)+C# After using #2x=3tanu#, #tanu=(2x)/3# and #u=arctan((2x)/3)# inverse transforms, I found #int dx/(4x^2+9)^2# =#1/108arctan((2x)/3)+1/216*(2*(2x)/3)/(((2x)/3)^2+1)+C# =#1/108arctan((2x)/3)+x/(72x^2+162)+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 5571 views around the world You can reuse this answer Creative Commons License