Compute the definite integral:
integral_0^∞ 1/(x^2 + 8 x + 20) dx
For the integrand 1/(x^2 + 8 x + 20), complete the square:
= integral_0^∞ 1/((x + 4)^2 + 4) dx
For the integrand 1/((x + 4)^2 + 4), substitute u = x + 4 and du = dx.
This gives a new lower bound u = 4 + 0 = 4 and upper bound u = ∞:
= integral_4^∞ 1/(u^2 + 4) du
Factor 4 from the denominator:
= integral_4^∞ 1/(4 (u^2/4 + 1)) du
Factor out constants:
= 1/4 integral_4^∞ 1/(u^2/4 + 1) du
For the integrand 1/(u^2/4 + 1), substitute s = u/2 and ds = 1/2 du.
This gives a new lower bound s = 4/2 = 2 and upper bound s = ∞:
= 1/2 integral_2^∞ 1/(s^2 + 1) ds
Apply the fundamental theorem of calculus.
The antiderivative of 1/(s^2 + 1) is tan^(-1)(s):
= lim_(b->∞) 1/2 tan^(-1)(s) right bracketing bar _2^b
Evaluate the antiderivative at the limits and subtract.
lim_(b->∞) 1/2 tan^(-1)(s) right bracketing bar _2^b = (lim_(b->∞) 1/2 tan^(-1)(b)) - 1/2 tan^(-1)(2):
= (lim_(b->∞) 1/2 tan^(-1)(b)) - 1/2 tan^(-1)(2)
lim_(b->∞) 1/2 tan^(-1)(b) = π/4:
Answer: |= 1/4 (π - 2 tan^(-1)(2))
