¡ì(up:1 down:0) (x^3 dx)/(x^2+2x+1)
x^3/(x+1)^2 = x - 2 + 3/(x+1) - 1/(x+1)^2
Now it's a piece of cake.
... Right?
at least it looks like the last one :)
To integrate the expression (∫(x^3 dx)) / (x^2 + 2x + 1), you can start by using algebraic manipulation to simplify the integrand.
First, you can factor the denominator (x^2 + 2x + 1) by recognizing that it is a perfect square trinomial. It can be factored as (x + 1)^2.
Now, let's rewrite the expression with the factored denominator:
∫(x^3 dx) / (x + 1)^2
To proceed further, you can use the method of substitution. Let's substitute u = x + 1, which means x = u - 1. Also, dx = du.
Rewriting the integral with the new variable and its derivative:
∫((u - 1)^3 du) / u^2
Now, you can expand the numerator and rewrite the integral as:
∫(u^3 - 3u^2 + 3u - 1) / u^2 du
Next, let's split the integral into separate terms:
∫(u^3/u^2) du - ∫(3u^2/u^2) du + ∫(3u/u^2) du - ∫(1/u^2) du
Simplifying further:
∫u du - 3 ∫du + 3 ∫(1/u) du - ∫(1/u^2) du
Now, integrate each term separately:
∫u du = (1/2)u^2 + C1
-3 ∫du = -3u + C2
3 ∫(1/u) du = 3 ln|u| + C3
- ∫(1/u^2) du = u^(-1) + C4
Now, substitute back u = x + 1 into the expressions:
(1/2)(x + 1)^2 - 3(x + 1) + 3 ln|x + 1| - (x + 1)^(-1) + C
This is the final result of the integration of (∫(x^3 dx) / (x^2 + 2x + 1))