Intergration and center of gravity

[tapout1829]

hey guys i need help with a problem i dont know where to start or what to do.

Thanks....



evaluate (INT x Tan^-1 x) dx
using intergration by parts with u = tan^-1 and v = (x^2 / 2) + (1/2)

and



center of gravity...

y= (-2x^2 / 169) + (4x / 13) + 3 (0<x<26) (< = less that or equal to)

actual distance the center of gravity travels is approx. 26.4046

evaluate the definite integral by first using the subsitution --> tan u = (-4 / 169)x + (4 / 13) and then integrate sec^3u


thanks alot any help with be greatly appreciated[/code]

 

[ChaoticLlama]

I'll start you off on the first one:

\(\displaystyle \L\int {x\arctan (x)dx}\)

let
\(\displaystyle \L\begin{array}{l}
u = \arctan (x) \\
du = \frac{{dx}}{{1 + x^2 }} \\
\end{array}
\L\begin{array}{l}
dv = xdx \\
v = \frac{{x^2 }}{2} \\
\end{array}\)

Therefore we have
\(\displaystyle \L\int {x\arctan (x)dx} = \frac{{x^2 \arctan (x)}}{2} - \int {\frac{{x^2 dx}}{{1 + x^2 }}}\)

Can you finish it from here?

 

[tapout1829]

Yeah i think i can thanks for the help
but i use as my v= (x^2/2) + (1/2) = (1/2)(x^2+1)

((x^2+1)arctan x / 2) - 1/2 INT dx


......== (1/ 2)(x^2+1)arctanx-x)+c