How do you integrate #1/(x^2+25)#? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Andrea S. Jun 27, 2018 #int dx/(x^2+25) = 1/5arctan(x/5) +C# Explanation: Substitute: #x=5t#, #dx = 5dt# to have: #int dx/(x^2+25) = int (5dt)/((5t)^2+25)# #int dx/(x^2+25) = 1/5 int dt/(t^2+1)# #int dx/(x^2+25) = 1/5arctan t +C# #int dx/(x^2+25) = 1/5arctan(x/5) +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 6318 views around the world You can reuse this answer Creative Commons License