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Thread: integration by parts

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    integration by parts

    Please can you help with the following:

    Q1)

    find dy/dx when y=(5x + 2)(3x - 1)

    Q2)

    with the help of integration by parts find:

    ∫ (2x+1) ln x dx

    many thanks for any help given!
    Luke.
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    with the help of integration by parts find:

    ∫ (2x+1) ln x dx

    many thanks for any help given!
    Luke.
    Let$\displaystyle u=ln(x), \;\ dv=(2x+1)dx, \;\ du=\frac{1}{x}dx, \;\ v=x^{2}+x$

    Put it together and it should work out fine.
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    Luke, If I were you, I would go to this website Wolfram|Alpha
    In the input window, type in this exact phrase: integrate integrate (2x+1)ln[x]dx. (you can copy & paste)
    Click the equals bar at the end if the input window.
    Be sure you click ‘show steps’ to see the solution.

    You can also use it to find: differentiate (5x^2 +2)^3 (3x-1)^2
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    Quote Originally Posted by luke.1985 View Post
    Please can you help with the following:

    Q1)

    find dy/dx when y=(5x + 2)(3x - 1)

    Q2)

    with the help of integration by parts find:

    ∫ (2x+1) ln x dx

    many thanks for any help given!
    Luke.
    Q1. You can either expand everything out, or use a combination of the chain and product rules.

    $\displaystyle y = (5x^2 + 2)^3(3x - 1)^2$

    $\displaystyle \frac{dy}{dx} = (5x^2 + 2)^3\,\frac{d}{dx}[(3x - 1)^2] + (3x - 1)^2\,\frac{d}{dx}[(5x^2 + 2)^3]$

    $\displaystyle = 6(3x - 1)(5x^2 + 2)^3 + 15(3x - 1)^2(5x^2 + 2)^2$

    $\displaystyle = 3(3x - 1)(5x^2 + 2)^2[2(5x^2 + 2) + 5(3x - 1)]$
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    Quote Originally Posted by luke.1985 View Post
    Please can you help with the following:

    Q1)

    find dy/dx when y=(5x + 2)(3x - 1)

    Q2)

    with the help of integration by parts find:

    ∫ (2x+1) ln x dx

    many thanks for any help given!
    Luke.
    Q2. $\displaystyle \int{(2x + 1)\ln{x}\,dx}$.

    Remember that $\displaystyle \int{u\,dv} = uv - \int{v\,du}$

    Let $\displaystyle u = \ln{x}$ so $\displaystyle du = \frac{1}{x}$

    Let $\displaystyle dv = 2x + 1$ so $\displaystyle v = x^2 + x = x(x + 1)$.


    So $\displaystyle \int{(2x + 1)\ln{x}\,dx} = x(x + 1)\ln{x} - \int{x(x + 1)\frac{1}{x}\,dx}$

    $\displaystyle = x(x + 1)\ln{x} - \int{x + 1\,dx}$

    $\displaystyle = x(x + 1)\ln{x} - x^2 - x + C$

    $\displaystyle = x(x + 1)(\ln{x} - 1) + C$
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