Calculus 2 correction

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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

Thanks

• Calculus 2 correction -

I have
u = x^2 + 2 so x^2 = (u-2)
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)(u-2)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

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