1. ## Two integration problems

Integrate 1/(x^3+x^2+x) dx I tried factoring the bottom and using the A/Factor1+B/Factor2 technique but couldn't really get anywhere with that.

Integrate 1/(1+cosx) dx I feel like this is simple but I can't figure it out.

2. Like you said, you would need to use partial fractions: $\displaystyle \frac{1}{x \left(x^2 + x + 1\right)} = \frac{A}{x} + \frac{Bx + C}{x^2 + x + 1}$

You can verify that: $\displaystyle A = 1, \ B = -1, \ C = -1$

So our new integral is: $\displaystyle \int \left( \frac{1}{x} - \frac{x + 1}{x^2+x+1}\right) dx$

For the second term, you're going to have to complete the square and use a trigonometric substitution. See how far you can go with that one. It'll probably be pretty long and tedious.

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For the second integral, multiply both top and bottom by $\displaystyle 1-\cos x$.

Now use a few trigonometric identities to get your integral to become: $\displaystyle \int \left( \csc^2 x - \csc x \cot x \right)\ dx$

which are standard ones.

3. Originally Posted by jimduquettesucked
Integrate 1/(x^3+x^2+x) dx I tried factoring the bottom and using the A/Factor1+B/Factor2 technique but couldn't really get anywhere with that.

Integrate 1/(1+cosx) dx I feel like this is simple but I can't figure it out.

$\displaystyle \int\frac{dx}{1+\cos(x)}=\int\frac{1-\cos(x)}{1-\cos^2(x)}=\int\frac{1-\cos(x)}{\sin^2(x)}=\int\left\{\csc^2(x)-\frac{\cos(x)}{\sin^2(x)}\right\}$