# Three integrals to evaluate

1. Sep 4, 2009

### clairez93

1. The problem statement, all variables and given/known data

1. $$\int\frac{2}{e^{-x} + 1}$$

$$\int\frac{2t - 1}{t^{2} + 4}$$

$$\int\frac{4}{4x^{2} + 4x + 65}$$

2. Relevant equations

3. The attempt at a solution

1. I'm not sure what to do. A u-substitution would give me e^(-x) dx, but I don't see how I could get an e^(-x) dx anywhere in the integrand.

2. Again I have no idea. I'm thinking maybe it will involve arc tan, except the numerator isn't 1, so what would I do with the 2t-1 numerator?

3. I tried completing the square, to get something that resembled an arc tan, but that didn't work too well. I'm not sure what else to do; a u-substitution wouldn't work I don't think that would just get 8x + 4.

No integration by parts or partial fractions allowed yet. Any help appreciated.

2. Sep 4, 2009

### zcd

For the first one, note that $$\int\frac{2}{e^{-x} + 1}\,dx=\int \frac{2}{1+\frac{1}{e^x}}}\,dx$$.

For the second one $$\int\frac{2t - 1}{t^{2} + 4}\,dt = \int\frac{2t}{t^{2} + 4}\,dt - \int\frac{1}{t^{2} + 4}\,dt$$

For the third one, you almost have it. Try doing 2x+1 = 8tanθ

3. Sep 4, 2009

### clairez93

Thanks for the other two hints, I don't understand what you mean by 2x+1 = tan theta?

4. Sep 4, 2009

### zcd

Are you familiar with trigonometric substitution?
$$\int\frac{4}{4x^{2} + 4x + 65}\,dx=\int\frac{4}{(2x+1)^{2}+8^{2}}\,dx=\frac{1}{16}\int\frac{1}{\tan^{2}(\theta)+1}\,dx$$

because
$$8\tan(\theta)=2x+1$$, $$\theta=\arctan(\frac{2x+1}{8})$$ and $$2dx=8\sec^{2}(\theta)$$.

5. Sep 4, 2009

### clairez93

Oooh nevermind I got it