# Integral and Arctangent

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

Mark44
Mentor
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
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.

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

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

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