General Antiderivative of (2x^3+3x^2+x-1)/((x+1)*((x^2+2x+2)^2))

[kallazans]

if you are not lazy, you will answer what is the general antiderivative of (2x^3+3x^2+x-1)/((x+1)*((x^2+2x+2)^2))

 

[suyver]

Originally posted by kallazans
S((2x^3+3x^2+x-1)/(x+1)(x^2+2x+2))dx!
are you like this?

I don't understand your question, and your equation is unclear. Do you mean

$$\int ((2x^3+3x^2+x-1)/(x+1)(x^2+2x+2)) {\rm d}x =$$

$$\int\frac{2x^3+3x^2+x-1}{x+1}(x^2+2x+2){\rm d}x =$$

$$-x+\frac{x^2}{2}+2x^3+\frac{5x^4}{4}+\frac{2x^5}{5}-\log(1+x)$$

or do you mean

$$\int ((2x^3+3x^2+x-1)/\left((x+1)(x^2+2x+2))\right) {\rm d}x =$$

$$\int\frac{2x^3+3x^2+x-1}{(x+1)(x^2+2x+2)}{\rm d}x =$$

$$2x-\arctan(1+x)-\log(1+x)-\log(2+(2+x)x)$$

 

[M98Ranger]

The general antiderivative of the given function can be found by using the partial fraction decomposition method. First, we need to factor the denominator into its irreducible factors, which are (x+1) and (x^2+2x+2)^2.

Next, we set up the partial fraction decomposition as follows:

(2x^3+3x^2+x-1)/((x+1)*((x^2+2x+2)^2)) = A/(x+1) + (Bx+C)/(x^2+2x+2) + (Dx+E)/(x^2+2x+2)^2

We can find the values of A, B, C, D, and E by equating the coefficients of each term on both sides of the equation. After solving for these values, we get:

A = -1/4, B = 1/2, C = 1/4, D = -1/8, E = -1/8

Therefore, the general antiderivative of the given function is:

-1/4 * ln(x+1) + 1/2 * ln(x^2+2x+2) + 1/4 * arctan(x+1) - 1/8 * (x+1)/(x^2+2x+2) - 1/8 * ln(x^2+2x+2) + C

where C is the constant of integration.