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# Help!!! Integration watch

1. If anyone can help me it will be most appreciated. I am having trouble with integrating the following equation. I do not know where to start... Can anyone give my a step by step solution? thanks.

y = (ln(x+1))^2

I am really stuck... thank you all for your time. Also does anyone know anything about Volume of revolution? My original equation is

y = ln(x+1)

at x = 0 to x = 6

I plotted this equation and am trying to find its volume under the curve by using volume of revolution. I think you have to square your original equation then integrate it. the equation is squared at the top and I am stuck. So if anyone could help that will be great. Thankyou.
2. (int from 0 to 6) (ln(x + 1))^2 dx
= (int from 1 to 7) (ln(u))^2 du
= [u (ln(u))^2] (from 1 to 7) - (int from 1 to 7) u(2 ln(u)/u) du
= 7(ln(7))^2 - 2(int from 1 to 7) ln(u) du
= 7(ln(7))^2 - 2[u ln(u) - u] (from 1 to 7)
= 7(ln(7))^2 - 14 ln(7) + 12

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