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How do I evaluate ∫ (x)/(3+5x) dx?

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How do I evaluate ∫ (x)/(3+5x) dx?

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  1. ∫ [x /(3 + 5x)] dx =

    it is a rational integrand whose numerator and denominator are of the same

    degree, whereas the degree of the numerator should be lower;

    therefore, in order to reduce its degree, do as follows:

    divide and multiply the integral by 5:

    (1/5) ∫ [5x /(3 + 5x)] dx =

    add (3 - 3) to the numerator:

    (1/5) ∫ [(5x + 3 - 3)/(3 + 5x)] dx =

    break it up into:

    (1/5) ∫ {[(5x + 3)/(3 + 5x)] - [3/(3 + 5x)]} dx =

    (1/5) ∫ [(5x + 3)/(3 + 5x)] dx - (1/5) ∫ [3/(3 + 5x)] dx =

    simplify the first integral:

    (1/5) ∫ dx - (1/5) ∫ [3/(3 + 5x)] dx =

    (1/5)x - (1/5) ∫ [3/(3 + 5x)] dx =

    as for the remaining integral, attempting to change the numerator into the

    derivative of the denominator, divide and multiply by (5/3):

    (1/5)x - (1/5)(3/5) ∫ {[(5/3)3]/(3 + 5x)} dx =

    (1/5)x - (3/25) ∫ [5/(3 + 5x)] dx =

    that is:

    (1/5)x - (3/25) ∫ [d(3 + 5x)]/ (3 + 5x) =

    (1/5)x - (3/25) ln | 3 + 5x | + C

    in conclusion:

    ∫ [x /(3 + 5x)] dx = (1/5)x - (3/25) ln | 3 + 5x | + C

    I hope it helps..

    Bye!


  2. ∫ (x)/(3+5x) dx

    let 3+5x=u

    5x=u-3

    x=(u-3)/5

    5 dx = du

    dx =(1/5) du

    The given integral becomes

    (1/5)(1/5) ∫ (u-3)du /u

    =(1/25)  ÃƒÂ¢Ã‚ˆÂ« [ 1-3/u] du

    =u/25 -(3/25) ln|u| +C

    =(3+5x) / 25 - (3/25) ln }3+5x| +C

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