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i. ∫ x²exdx
 u = x²    dv = exdx
du = 2xdx   v = ex
uv - ∫vdu
(x²)ex - ∫ex(2xdx)
(1)     x²ex - 2∫xexdx
Now we do ∫xexdx 
 u = x     dv = exdx
du = dx    v = ex
uv - ∫vdu   
 
xex - ∫exdx
xex - ex
Now substitute that for the integral in expression (1),
and put on the arbitrary constant + C
(1)     x²ex - 2∫xexdx
        x²ex - 2(xex - ex)
        x²ex - 2xex + 2ex + C
ii. ∫ x ln(x)dx
 u = ln(x)     dv = xdx
du =  dx       v =
dx       v =  x²
You finish.
Edwin
x²
You finish.
Edwin