How do you integrate 1x2+25? Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Andrea S. Jun 27, 2018 ∫dxx2+25=15arctan(x5)+C Explanation: Substitute: x=5t, dx=5dt to have: ∫dxx2+25=∫5dt(5t)2+25 ∫dxx2+25=15∫dtt2+1 ∫dxx2+25=15arctant+C ∫dxx2+25=15arctan(x5)+C Answer link Related questions How do I evaluate the indefinite integral ∫sin3(x)⋅cos2(x)dx ? How do I evaluate the indefinite integral ∫sin6(x)⋅cos3(x)dx ? How do I evaluate the indefinite integral ∫cos5(x)dx ? How do I evaluate the indefinite integral ∫sin2(2t)dt ? How do I evaluate the indefinite integral ∫(1+cos(x))2dx ? How do I evaluate the indefinite integral ∫sec2(x)⋅tan(x)dx ? How do I evaluate the indefinite integral ∫cot5(x)⋅sin4(x)dx ? How do I evaluate the indefinite integral ∫tan2(x)dx ? How do I evaluate the indefinite integral ∫(tan2(x)+tan4(x))2dx ? How do I evaluate the indefinite integral ∫x⋅sin(x)⋅tan(x)dx ? See all questions in Integrals of Trigonometric Functions Impact of this question 6950 views around the world You can reuse this answer Creative Commons License