How do you find the integral of #int 1/(3+(x-2)^2#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Ratnaker Mehta Jul 8, 2017 # 1/sqrt3*arc tan((x-2)/sqrt3)+C.# Explanation: Let, #I=int1/(3+(x-2)^2)dx.# Knowing that, #int1/(t^2+a^2)dt=1/a*arc tan(t/a)+c,# we take substn. #(x-2)=t rArr dx=dt.# #:. I=int1/(t^2+sqrt3^2)dt,# #=1/sqrt3*arc tan (t/sqrt3),# # rArr I=1/sqrt3*arc tan((x-2)/sqrt3)+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 1346 views around the world You can reuse this answer Creative Commons License