Compute the Integral

[quex]

Compute the integral:

2
∫ arctan(x) dx/ (x^2 - x +1)
1/2


Answer is (π^2√3)/18


I cannot figure this one out. My teacher posted the solution but I cannot follow its steps. I personally believe there is a typo on it but I am not certain.


Solution:

The solution involves the averaging of 2 integrals.


We know: arctan(x) + arctan(1/x) = π/2


Thus, let y = 1/x: (Note: the definite integral bounds are from 1/2 to 2 but are not shown)

(1/2) (∫ arctan(x) dx / (x^2 - x +1) + ∫ (π/2 - arctan(y)) dy / (y^2 - y +1) ).


We find that dx/(x^2 - x +1) = - dy/(y^2 - y +1).


The average of the integrals comes out to be (π/4)∫ dx/(x^2 - x +1).
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Solving the last integral renders the solution. However whenever I average the two listed integrals, I keep on getting zero. This is what my teacher's solution says. I personally don't understand why those two integrals were added because they seem arbitrary.

Thank you in advance! This is a tough problem.

 

[JohnZ]

2
∫ arctan(x) dx/ (x^2 - x +1)
1/2

2
∫ arctan(x) dx/ (x^2 - x +1) =
1/2
2.....1
∫ + ∫ ............... (1)
1....1/2

for 2nd integral, use substitution x = 1/y, dx = -1/y^2 dy, get
2
∫ arctan(1/y) dy / (y^2 -y +1)
1

use arctan(x) + arctan(1/x) = pi/2

(1) become pi/2 *
2
∫ dx / (x^2 - x + 1)
1

2
∫ dx / (x^2 - x + 1) = pi * √3 / 9
1

(hint: complete square and do some substitution.., you'll get C*
√3
∫ du / (u^2 + 1), C = 2/√3 )
1/√3

sorry don't know how to write math equation.