Integral check ∫ 6 x + 4 x 4 + 3 x 2 + 5 dx

Question

Integral check   ∫            6      x      +      4                      x        4            +      3              x        2            +      5           dx

Cagliusov8 · Accepted Answer

All the signs of all your letters are switched. This is because

A + A + C = A + 0 = A = 6

, etc. Also,

(x^{2} + 1) (x^{2} + 2) = x^{4} + 3 x^{3} + 2 \neq x^{2} + 3 x + 5

Your polynomial has no real roots, so you'll have to do some manipulation in

C

, as follows:
Let

x^{2} = u

. Then

u^{2} + 3 u + 5 = 0

has solutions

u_{1} = \frac{- 3 + i \sqrt{1} 1}{2}

u_{2} = \frac{- 3 - i \sqrt{1} 1}{2}

Thus

u^{2} + 3 u + 5 = (u - \frac{- 3 - i \sqrt{1} 1}{2}) (u - \frac{- 3 + i \sqrt{1} 1}{2})

We can write then

u^{2} + 3 u + 5 = (u + \frac{3}{2} + \frac{i \sqrt{1} 1}{2}) (u + \frac{3}{2} - \frac{i \sqrt{1} 1}{2})

and using the difference of squares factorization

u^{2} + 3 u + 5 = {(u + \frac{3}{2})}^{2} - \frac{i^{2} 11}{4} = {(u + \frac{3}{2})}^{2} + \frac{11}{4}

Thus

x^{4} + 3 x^{2} + 5 = {(x^{2} + \frac{3}{2})}^{2} + \frac{11}{4}

Integral check int (6x+4)/(x^4+3x^2+5)dx

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