Integral and Arctangent

1. Oct 1, 2014

norman95

What did I do wrong evaluating this integral: ∫ [4 / (4x^2 + 4x + 65) dx ?

Here were my steps:
1) = 4 ∫ [1 / (4x^2 + 4x + 65)] dx
2) = 4 ∫ [1 / (4x^2 + 4x + 1 + 64)] dx
3) = 4 ∫ [1 / ((2x+1)^2 + 64)] dx ;

Then I applied the formula: ∫ [1 / √(a^2 + u^2)] dx = (1/a)arctan(u/a) + C;

4) = (4)(1/8)arctan [(2x+1) / (√64)] + C
5) = (1/2)arctan [(2x+1) / 8] + C

My answer: (1/2)arctan [(2x+1) / 8] + C

The correct answer: (1/4)arctan[(2x+1) / 8] + C

2. Oct 1, 2014

Staff: Mentor

Your answer is too large by a factor of 2. I suspect that when you made the substitution from x to u, you forgot to change from dx to du correctly. If u = 2x + 1, then du = 2*dx.

BTW, when you do an integration problem, it is a good idea to check your work by differentiating your answer. You should end up with the same integrand you started with.

3. Oct 2, 2014

HallsofIvy

Staff Emeritus
There is no such formula. Did you mean "$\int \frac{1}{a^2+ u^2} du= arctan(u/a)+ C$"?

Last edited: Oct 2, 2014