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

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

2 Answers
Andrea S.
Jun 5, 2018

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

Steve M
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 #

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