Would like to solve this integral but it kind of messy especially when it comes to end?

int x/(x^2 +x+1)dxxx2+x+1dx please include explanations on where you do the substitutions and all factoring

2 Answers
Jun 5, 2018

intx/(x^2+x+1)dx = 1/2ln(x^2+x+1)-1/sqrt3arctan((2x+1)/sqrt3)+Cxx2+x+1dx=12ln(x2+x+1)13arctan(2x+13)+C

Explanation:

First we examine the denominator: as the determinant:

Delta = 1-4= -3 <0

is negative, the denominator cannot be factorized.

So, we split the integrand, by obtaining at the numerator the derivative of the denominator, and compensating:

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

Using the linearity of the integral:

intx/(x^2+x+1)dx = 1/2int (2x+1)/(x^2+x+1)dx -1/2int dx/(x^2+x+1)

Now the first integral can be solved directly:

int (2x+1)/(x^2+x+1)dx = int (d(x^2+x+1))/(x^2+x+1) = ln(x^2+x+1)+C

while for the second we complete the square at the denominator:

int dx/(x^2+x+1) = int dx/((x+1/2)^2+3/4)

int dx/(x^2+x+1) = 4 int dx/((2x+1)^2+3)

int dx/(x^2+x+1) = 4/3 int dx/(((2x+1)/sqrt3)^2+1)

int dx/(x^2+x+1) = 2/sqrt3 int (d((2x+1)/sqrt3))/(((2x+1)/sqrt3)^2+1)

int dx/(x^2+x+1) = 2/sqrt3 arctan((2x+1)/sqrt3)+C

Putting the partial results together:

intx/(x^2+x+1)dx = 1/2ln(x^2+x+1)-1/sqrt3arctan((2x+1)/sqrt3)+C

Jun 5, 2018

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, I_1, we can perform a substitution. Let:

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_3, we can complete the square on the denominator:

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 x^2+x+1 gt AA x In RR

I = 1/2 ln (x^2+x+1) -sqrt(3)/3 arctan((2x+1)/sqrt(3)) + C