- #1
norman95
- 6
- 1
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
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