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# No idea!! watch

1. can someone help me in this prom

how to integrate ln(1+x)
2. 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
4. No prob.
5. Damn! This is what I have to look forward to for Edexcel P3 eh? Well, I am dead.

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Updated: July 11, 2004
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