How do you Integrate ln(x^2 +4) or x*sqrt(3-x)? I am stuck, any sort of help will be appreciated, thanks! =)?

Question

I have tried already to do it myself, with absolutely NO results =( Desperate for an answer now...thanks!

Answer 1

Use integartion by parts
Let u=ln(x^2+4), dv/dx=1
du/dx=2x/(x^2+4), v=x
int(udv/dx)=uv-int(vdu/dx)
=xln(x^2+4)-int(2x^2/(x^2+4)) simplify by long division
=xln(x^2+4)-int(2-8/(x^2+4))
=xln(x^2+4)-(2x-4arctan(x/2))

Answer 2

Use integration by parts for x*sqrt(3-x) where

u=x and dv=(3-x)^(1/2)

As for the ln(x^2+4) I am not sure what to do with that one.

Answer 3

Integrate using by parts method as suggested by z_o_r_r_o.

Write ln(x^2 +4) as1*ln(x^2 +4). Take 1 as dv and proceed as you do in integration by parts.