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Thread: Integrals! help!

  1. #1
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    Integrals! help!

    need to get the integral of:

    3x - x/((x^2+2)^4/3)

    substitution rule here? or is it ln. i'm not sure
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  2. #2
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    Hello,

    You can integer 3x on its own.

    For the second term, you can rewrite it into $\displaystyle - \frac{1}{2} 2x (x^2+2)^{-\frac{4}{3}} = \frac{1}{2} 3* 2x (x^2+2)^{-\frac{4}{3}}$

    $\displaystyle -3$ is $\displaystyle \frac{1}{-\frac{4}{3} + 1}$, 2x is the derivate of x

    With all that elements you should be able to find the integral, using only formulas you learnt
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    i dont understand that at all.. i'm new to integrals.. can you show me how to do it directly sorry
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    Quote Originally Posted by s0urgrapes View Post
    need to get the integral of:

    3x - x/((x^2+2)^4/3)

    substitution rule here? or is it ln. i'm not sure
    To integrate $\displaystyle \frac{x}{(x^2 + 2)^{4/3}}$, make the substitution $\displaystyle u = x^2 + 2$. Note that $\displaystyle \frac{du}{dx} = 2x \Rightarrow dx = \frac{du}{2x}$.

    So $\displaystyle \int \frac{x}{(x^2 + 2)^{4/3}} \, dx = \int \frac{x}{u^{4/3}} \, \frac{du}{2x} = \frac{1}{2} \, \int \frac{1}{u^{4/3}} \, du = \frac{1}{2} \, \int u^{-4/3} \, du = ....$

    Remember to substitute back $\displaystyle u = x^2 + 2$ at the finish ....
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  5. #5
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    Ok...

    First of all, note the formulas you should know :

    The integral of $\displaystyle u'(x) u^n(x)$ is $\displaystyle \frac{1}{n+1} u^{n+1}(x)$

    We have here $\displaystyle 3x - \frac{x}{(x^2 + 2)^{4/3}} = 3x - \frac{1}{2} 2x (x^2+2)^{-4/3}$

    The integral of 3x is 3x/2

    In the second term, u(x)=x+2. u'(x)=2x.
    n=-4/3, so n+1 will be -1/3 and 1/(n+1) will be -3

    Hence the integral is :

    $\displaystyle \frac{3x^2}{2} - \frac{1}{2} (x^2+2)^{-1/3} * (-3) = \frac{3x^2}{2} + \frac{3}{2} \frac{1}{(x^2+2)^{-1/3}} = \frac{3}{2} (x^2 + \frac{1}{(x^2+2)^{-1/3}})$
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  6. #6
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    thanks!
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