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    • Thread Starter
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    can someone help me in this prom

    how to integrate ln(1+x)
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    Do it by parts:

    u = ln(x + 1) => du/dx = 1/(x + 1)
    dv/dx = 1 => v = x

    So, INT ln(x + 1) dx = uv - INT v du/dx

    = xln (x + 1) - INT x dx/(x + 1)

    To do the integral of x dx/(x + 1), use a trick: rewrite the top as (x + 1) - 1.

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

    = 1 - 1/(x + 1)

    So, INT x dx/(x + 1) = INT 1 - 1/(x + 1) dx

    = x - ln (x + 1)

    => INT ln (x + 1) = xln (x + 1) - [x - ln(x + 1)] + C

    = xln (x + 1) - x + ln (x + 1) + C
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    Thank your dude!
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    No prob.
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    Damn! This is what I have to look forward to for Edexcel P3 eh? Well, I am dead.
 
 
 
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