Problem 1.

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

[tex]\int[/tex][tex]\frac{dx}{1+cos^2(x)}[/tex]

**2. The attempt at a solution**

I've tried writing it in these ways:

[tex]\int\frac{dx}{2-sin^2(x)}[/tex]

[tex]\int\frac{dx}{(secx+cosx)(cosx)}[/tex]

Let u=cosx

[tex]\int\frac{-du}{\sqrt{1-u^2}(1+u^2)}[/tex]

So far I've gotten nowhere. I've also tried to make the integral look like the integral for arctanx. I realize that I need to have a -sinx in the numerator but don't know how to make it work out.

Problem 2:

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

[tex]\int[/tex][tex]\frac{dx}{sec^2(x)+tan^2(x)}[/tex]

**2. The attempt at a solution**

I've tried writing it like:

[tex]\int[/tex][tex]\frac{cos^2(x)}{1+sin^2(x)}[/tex]

U-substitution and trig-substition don't seem to work.

Problem 3:

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

[tex]\int[/tex][tex]\frac{dx}{x^3+1}[/tex]

**2. The attempt at a solution**

In this problem I actually got somewhere:

I split the denominator into two polynomials and used partial fractions.

I ended with:

[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{3}[/tex][tex]\int[/tex][tex]\frac{(x-2)dx}{x^2-x+1}[/tex]

Then:

[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex][tex]\int[/tex][tex]\frac{(2x-4)dx}{x^2-x+1}[/tex]

[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex][tex]\int[/tex][tex]\frac{(2x-1)dx}{x^2-x+1}[/tex] - [tex]\int[/tex][tex]\frac{-3dx}{x^2-x+1}[/tex]

[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex]ln(x^2-x+1)+3[tex]\int[/tex][tex]\frac{dx}{x^2-x+1}[/tex]

Now I'm stuck.... don't know what to do.

Any help would be greatly appreciated!