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Math Help - Integrations by parts

  1. #1
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    Integrations by parts

    Intergrate these:

    (x^2) dx
    (2+3x)^1/2

    and

    (x^2-25)^1/2 dx
    x


    We have to intergrate them using parts and it's so frustrating because i just don't know how to do it! please help, thank you!
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  2. #2
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    \int \frac{x^2}{\sqrt{3x + 2}} ~dx = \int x^2 ~(3x + 2)^{-\frac{1}{2}} ~dx.

    Remember the rule for integration by parts: \int u~dv = u~v - \int v~du

    Letting u = x^2 and dv = (3x + 2)^{-\frac{1}{2}}, so du = 2x and v = \frac{2}{3} \sqrt{3x + 2}

    = \frac{2}{3}x^2~\sqrt{3x + 2} - \int \frac{4}{3}x~\sqrt{3x + 2}~dx

    Now we will leave the first part of this equation alone just for now and deal with the second part which requires another integration by parts.

    Let u = x and v = \sqrt{3x + 2}. So du = 1 and v = \frac{2}{9} (3x + 2)^{\frac{3}{2}}

    \frac{4}{3}~\int x~\sqrt{3x + 2}~dx = \frac{4x}{3} \times\frac{2}{9} (3x + 2)^{\frac{3}{2}} - \frac{4}{3}\int~\frac{2}{9} (3x + 2)^{\frac{3}{2}} ~dx

    = \frac{8x}{27} (3x + 2)^{\frac{3}{2}} - \frac{8}{27} \times \frac{(3x + 2)^{\frac{5}{2}}}{\frac{5}{2} \times 3}

    = \frac{8x}{27} (3x + 2)^{\frac{3}{2}} - \frac{16}{405} (3x + 2)^{\frac{5}{2}}


    Now, returning to the equation we left off from, the solution to this problem is:

    = \frac{2}{3}x^2~\sqrt{3x + 2} - \int \frac{4}{3}x~\sqrt{3x + 2}~dx

    = \frac{2}{3}x^2~\sqrt{3x + 2} - \frac{8x}{27} (3x + 2)^{\frac{3}{2}} + \frac{16}{405} (3x + 2)^{\frac{5}{2}}

    If needed it can be simplified:

    = \frac{2}{3}\sqrt{3x + 2}~\left[x^2 - \frac{4x}{9} (3x + 2) + \frac{8}{135} (3x + 2)^2\right]

    = \frac{2}{405}\sqrt{3x + 2}~\left[135x^2 - 60x (3x + 2) + 8 (3x + 2)^2\right]

    = \frac{2}{405}\sqrt{3x + 2}~(27x^2 - 24x + 32)

    Now you try do the other one...
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  3. #3
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    thanks i get it now!
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