Calculus 2 correction
posted by Tom .
I just wanted to see if my answer if correct
the integral is:
∫(7x^3 + 2x  3) / (x^2 + 2)
when I do a polynomial division I get:
∫ 7x ((12x  3)/(x^2 + 2)) dx
so then I use
u = x^2 + 2
du = 2x dx
1/2 du = x dx
= ∫7x  12/2∫du/u  3/1∫1/u
= (7x^2)/2  6/u  3lnu + C
= (7x^2)/2  6/(x^2+2)  3ln(x^2+2) + c
is my answer good?
Thanks

I have
u = x^2 + 2 so x^2 = (u2)
du = 2 x dx so x dx =(1/2) du
split into three integrals
7 x^3 dx/u + 2 x dx/u  3 dx/(x^2+2)
7x^2 xdx/u+du/u(3/sqrt2)tan^1(x/sqrt2)
(7/2)(u2)du/u +du/u  last term
(7/2)du 7du/u+du/u  last term
(7/2)u  6du/u  last term
(7/2)(x^2+2)  6ln u  last term
(7/2)(x^2)  6 ln(x^2+2) last term + constant