Evaluate the following integral

Question

∫1/(4+t²)dt

Answer 1

∫ [1 / (4 + t²)]dt = ∫ [1 / (4(1 + (t/2)²)]dt

Let u = t/2, then du = (1/2)dt. So, dt = 2du.

We have (1/2) ∫ [ 1 / /(1 + u²) ]du = (1/2)Arctanu + C = (1/2)Arctan(t/2) + C

Answer 2

Multiply the numerator and denominator by 1/4.

Then the antiderivative will be an arctan function.