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Math Help - Two integration problems

  1. #1
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    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.

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

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

    So our new integral is: \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.

    _________________________

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

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

    which are standard ones.
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  3. #3
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    Quote Originally Posted by jimduquettesucked View Post
    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.

    Thanks for your help.
    \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\}
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