How to do the integration by parts of 1/(x(x+1))dx?

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  • MyRank
    Lv 6
    1 month ago

    1/(x(x+1))

    1/x(x+1) = A/x + B/(x+1)

    1 = A(x+1) + Bx

    Put x = 0

    1 = A(0 + 1) + B(0)

    1 = A

    Put x = -1

    1 = A(-1 + 1) + B(-1)

    1 = -B

    B=-1

    1/x(x+1) = 1/x - 1/(x+1)

    Integration on both sides

    ∫1/x(x+1). dx =∫1/x.dx - ∫1/(x+1).dx

    = log|x| - log|x+1| + logc

    = log|(x/(x+1)|c

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  • 1 month ago

    Yet another way, besides partial fractions, would be to rewrite the integrand as

    dx/[(x+1/2)^2 - (1/4)], then set

    x + 1/2 = (1/2)sec(u),

    so dx = (1/2)sec(u)tan(u) du, and

    (x+1/2)^2 - (1/4) = (1/4)[sec^2(u) - 1] = (1/4)tan^2(u).

    Now the integrand becomes

    (1/2)sec(u)tan(u) du/[(1/4)tan^2(u)]

    = 2 sec(u) du/tan(u) = 2 csc(u) du,

    and after integration you have

    -2 ln|csc(u) + cot(u)| + C.

    To get back to x, draw a triangle where u is the angle,

    x+1/2 is the hypotenuse, and (1/2) is the "adjacent" side.

    Then the opposite side is sqrt(x^2 + x), so the above logarithmic expression can be rewritten in "x".

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  • You don't integrate this by parts.  You use partial fraction decomposition

    1 / (x * (x + 1)) = a/x + b/(x + 1)

    a/x + b/(x + 1) =>

    a * (x + 1) / (x * (x + 1))  +  b * x / (x * (x + 1)) =>

    (a * (x + 1) + b * x) / (x * (x + 1)) =>

    (ax + bx + a) / (x * (x + 1)) =>

    ((a + b) * x + a) / (x * (x + 1))

    0x + 1 = (a + b) * x + a

    a + b = 0

    a = 1

    1 + b = 0

    b = -1

    Now you have:

    1 / x  -  1 / (x + 1)

    dx / x  -  dx / (x + 1)

    Integrate

    ln|x| - ln|x + 1| + C =>

    ln|x / (x + 1)| + C

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  • ted s
    Lv 7
    1 month ago

    you don't !...partial fraction decomposition...1/x - 1 / (x + 1 )

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