Would like to solve this integral but it kind of messy especially when it comes to end?
int x/(x^2 +x+1)dx∫xx2+x+1dx please include explanations on where you do the substitutions and all factoring
2 Answers
Explanation:
First we examine the denominator: as the determinant:
is negative, the denominator cannot be factorized.
So, we split the integrand, by obtaining at the numerator the derivative of the denominator, and compensating:
Using the linearity of the integral:
Now the first integral can be solved directly:
while for the second we complete the square at the denominator:
Putting the partial results together:
int \ x/(x^2 +x+1) \ dx = 1/2 ln (x^2+x+1) -sqrt(3)/3 arctan((2x+1)/sqrt(3)) + C
Explanation:
We seek:
I = int \ x/(x^2 +x+1) \ dx
We first manipulate the numerator such that we have the derivative of the denominator:
I = int \ (1/2(2x+1)-1/2)/(x^2 +x+1) \ dx
\ \ = 1/2 \ int \ (2x+1)/(x^2 +x+1) \ dx - 1/2 \ int \ 1 /(x^2 +x+1) \ dx
\ \ = 1/2 \ I_1 -1/2 I_2 , say,
Where:
I_1 = int \ (2x+1)/(x^2 +x+1) \ dx ,I_2= int \ 1 /(x^2 +x+1) \ dx
For the first integral,
u = x^2+x+1 => (du)/dx = 2x+1
And if we substitute into the integral, we get:
I_1 = int \ q/u \ du
\ \ \ = ln |u| + C
\ \ \ = ln |x^2+x+1| + C
Amd, for the first integral,
I_2 = int \ 1 /((x+1/2)^2-(1/2)^2+1) \ dx
\ \ \ = int \ 1 /((x+1/2)^2 + 3/4) \ dx
And, we can perform a substitution. Let:
x+1/2 = sqrt(3)/2u iff u=(2x+1)/sqrt(3)=> (dx)/(du) = sqrt(3)/2
And if we substitute into the integral, we get:
I_2 = int \ 1 /(3/4u^2 + 3/4) \ sqrt(3)/2 \ du
\ \ \ = 4/3 \ sqrt(3)/2 \ int \ 1 /(u^2 + 1) \ du
\ \ \ = 2/3 \ sqrt(3) arctanu + C
\ \ \ = 2/3 \ sqrt(3) arctan((2x+1)/sqrt(3)) + C
And, combining these results we have:
I = 1/2 \ I_1 -1/2 I_2
\ \ = 1/2 {ln |x^2+x+1|} -1/2 {2/3 \ sqrt(3) arctan((2x+1)/sqrt(3))} + C
\ \ = 1/2 ln |x^2+x+1| -sqrt(3)/3 arctan((2x+1)/sqrt(3)) + C
Finally, noting that
I = 1/2 ln (x^2+x+1) -sqrt(3)/3 arctan((2x+1)/sqrt(3)) + C