How do you find the integral of #(arctan(2x)) / (1+4x^2)#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Bill K. May 19, 2015 Do a substitution: #u=arctan(2x), du=1/(1+(2x)^2)\cdot 2\ dx=2/(1+4x^2)\ dx#, giving #\int arctan(2x)/(1+4x^2)\ dx=\frac{1}{2}\int u\ du=\frac{1}{4}u^{2}+C# Hence, #\int arctan(2x)/(1+4x^2)\ dx=\frac{1}{4}arctan^{2}(2x)+C#. Answer link Related questions How do I evaluate the indefinite integral #intsin^3(x)*cos^2(x)dx# ? How do I evaluate the indefinite integral #intsin^6(x)*cos^3(x)dx# ? How do I evaluate the indefinite integral #intcos^5(x)dx# ? How do I evaluate the indefinite integral #intsin^2(2t)dt# ? How do I evaluate the indefinite integral #int(1+cos(x))^2dx# ? How do I evaluate the indefinite integral #intsec^2(x)*tan(x)dx# ? How do I evaluate the indefinite integral #intcot^5(x)*sin^4(x)dx# ? How do I evaluate the indefinite integral #inttan^2(x)dx# ? How do I evaluate the indefinite integral #int(tan^2(x)+tan^4(x))^2dx# ? How do I evaluate the indefinite integral #intx*sin(x)*tan(x)dx# ? See all questions in Integrals of Trigonometric Functions Impact of this question 7086 views around the world You can reuse this answer Creative Commons License