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Math Help - calculus BC

  1. #1
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    Exclamation calculus BC

    i've been staring at these problems for at least an hour and still don't know where to start
    can anyone give me some hints?

    1.) integrate [(x^2)/(2+3x)^(1/2)] dx
    2.) integrate [(x^2-25)^(1/2)/(x)] dx
    i think this problem might be arcsin?? i don't know

    these problems are either integration by parts (q*r-Sr*dq) or right triangles (trig)

    thank you so much for any help!
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  2. #2
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    After staring for an hour, you should take a break.

    For the first, try u^{2} = 2+3x. It's not too bad. No Arcsine or anything past algebra.

    For the second, perhaps u^{2} = x^{2} - 25 will lead somewhere useful.
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  3. #3
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    yeah that's what i thought initially, but we are supposed to use integration by parts or right triangles
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  4. #4
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    Quote Originally Posted by holly123 View Post
    i've been staring at these problems for at least an hour and still don't know where to start
    can anyone give me some hints?

    1.) integrate [(x^2)/(2+3x)^(1/2)] dx
    2.) integrate [(x^2-25)^(1/2)/(x)] dx
    i think this problem might be arcsin?? i don't know

    these problems are either integration by parts (q*r-Sr*dq) or right triangles (trig)

    thank you so much for any help!
    (2) Try letting x = 5 \sec \theta \Rightarrow \frac{1}{25}x^2-1=\frac{1}{25}25 \sec^2 \theta-1  = \tan^2 \theta

    Then \int \frac{\sqrt{x^2-25}}{x}=\frac{\sqrt{25(\frac{1}{25}x^2-1)}}{x}=...
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  5. #5
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    Quote Originally Posted by holly123 View Post
    yeah that's what i thought initially, but we are supposed to use integration by parts or right triangles
    Who say's you won't get around to that after the initial substitution?
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