Fine the integra (3x+2/(x^2+x+1))dx?

To find the integral of (3x + 2)/(x^2 + x + 1) dx, we can first rewrite the expression as:

(3x + 2)/(x^2 + x + 1) = (3x + 2)/(x^2 + x + 1) = (3x + 2)/(x^2 + x + 1)

Next, we can use partial fraction decomposition to break down the expression into simpler fractions. Let's assume that the expression can be written as:

(3x + 2)/(x^2 + x + 1) = A/(x + 1) + B/(x^2 + 1)

Multiplying both sides by (x^2 + x + 1) to clear the denominators, we get:

3x + 2 = A(x^2 + 1) + B(x + 1)

Expanding and equating coefficients, we get:

3 = A
2 = A + B

Solving these equations, we find that A = 3 and B = -1. Therefore, the expression can be rewritten as:

(3x + 2)/(x^2 + x + 1) = 3/(x + 1) - 1/(x^2 + 1)

Now, we can integrate each term separately:

∫(3/(x + 1) - 1/(x^2 + 1)) dx = 3ln|x + 1| - arctan(x) + C

Therefore, the integral of (3x + 2)/(x^2 + x + 1) dx is:

3ln|x + 1| - arctan(x) + C

∫ (3x^2/(x^2 + 1)) dx

= ∫ (3x^2 + 3 - 3/(x^2 + 1)) dx

= ∫ (3x^2 + 3) dx - ∫ (3/(x^2 + 1)) dx

= x^3 + 3x - 3 arctan(x) + C