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Thread: Integration Help

  1. #1
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    Integration Help

    Hey guys, I'm new to integration and these are giving me a really rough time, even just a push in the right direction would be great. Thanks for taking a look! So far, I've only learned the basic integration techniques and u-substitution.

    1) $\displaystyle \int\frac{x+1}{\sqrt{x^2+2x-3}}dx$

    2) $\displaystyle \int\frac{dx}{x^{\frac{3}{4}}(x^\frac{1}{4}+2)^2}$

    I solved 2) and got $\displaystyle 16\sqrt[4]{x}-\frac{4}{\sqrt[4]{x}}$. Not sure if this is right, I did expanded completely and brought everything up top before integrating.

    3) $\displaystyle \int3x\sin(4-x^2)dx$

    4) $\displaystyle \int\frac{dx}{\cos^2x\sqrt{1+\tan{x}}}$

    Thanks for any pointers!
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  2. #2
    MHF Contributor arbolis's Avatar
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    For the first one, I'd try a $\displaystyle u$-substitution as $\displaystyle u=x^2+2x-3 \Rightarrow du=2x+2 dx$.
    Hence the integral is worth $\displaystyle \frac{1}{2} \int \frac{du}{\sqrt u}$. Finish it.
    Edit: I also see a u-sub for the third integral : let $\displaystyle u=4-x^2$... can you finish it?
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  3. #3
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    Quote Originally Posted by arbolis View Post
    For the first one, I'd try a $\displaystyle u$-substitution as $\displaystyle u=x^2+2x-3 \Rightarrow du=2x+2 dx$.
    Hence the integral is worth $\displaystyle \frac{1}{2} \int \frac{du}{\sqrt u}$. Finish it.
    Edit: I also see a u-sub for the third integral : let $\displaystyle u=4-x^2$... can you finish it?
    Wicked I think I worked out a solution for both of them, if you have a sec would you mind checking over my work?

    1) $\displaystyle \int\frac{x+1}{\sqrt{x^2+2x-3}}dx$

    Let $\displaystyle u=x^2+2x-3$

    $\displaystyle du=2x+2dx$

    $\displaystyle \frac{1}{2}du=(x+1)dx$

    Therefore

    $\displaystyle \frac{1}{2}\int\frac{du}{\sqrt{u}}$

    $\displaystyle =\frac{1}{2}\int u^{-{\frac{1}{2}}}du$

    $\displaystyle =(\frac{1}{2})(2u^{\frac{1}{2}})+c$

    $\displaystyle =\sqrt{u}+c$

    $\displaystyle \boxed{=\sqrt{x^2+2x-3}+c}$

    ----------------------------------------------------------
    And for number 3) I got the following;
    $\displaystyle
    \int3x\sin{(4-x^2)}dx$

    Let $\displaystyle u=4-x^2$

    $\displaystyle du=-2xdx$

    $\displaystyle -\frac{3}{2}du=3xdx$

    Therefore

    $\displaystyle \int(\sin{(u)})(-\frac{3}{2}du)$

    $\displaystyle =-\frac{3}{2}\int\sin(u)du$
    $\displaystyle
    =-\frac{3}{2}(-\cos{u})+C$
    $\displaystyle
    \boxed{=\frac{3}{2}\cos{(4-x^2)}+C}$

    What do you think?
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  4. #4
    MHF Contributor arbolis's Avatar
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    Both looks right to me. Very good!
    I'll let others helping you for the 2 remaining ones.
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  5. #5
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    Cool, thanks!
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