Calculus: Partial Fractions?

Question

For this problem I tried completing the square and then used trig substitution, but it doesn’t seem right because it forced me to have the lower limit of integration be arctan(2/3), which does not have one of the known exact values.

1. x dx / [x^2 + 4x + 13] (from 0 to 1; definite integral)

(which reads as x in the numerator, dived by x squared plus 4x plus 13 in the denominator).

Answer 1

Integrate ∫[x/(x² + 4x + 13)]dx from 0 to 1.

∫[x/(x² + 4x + 13)]dx
Let
3tanθ = x + 2
3sec²θ dθ = dx

x² + 4x + 13 = (x² + 4x + 4) + 9 = (x + 2)² + 9 = 9tan²θ + 9
x = 3tanθ - 2
x dx = (3tanθ - 2)(3sec²θ dθ)
___________

∫[x/(x² + 4x + 13)]dx

= ∫[(3tanθ - 2)(3sec²θ) / (9tan²θ + 9)]dθ

= ∫[(3tanθ - 2)(3sec²θ) / (9sec²θ)]dθ

= ∫[(3tanθ - 2) / 3]dθ

= ∫[tanθ - 2/3]dθ

= -ln|cosθ| - 2θ/3
______________

Now work up the new limits.

When x = 0
3tanθ - 2 = 0
3tanθ = 2
tanθ = 2/3
θ = arctan(2/3)

When x = 1
3tanθ - 2 = 1
3tanθ = 3
tanθ = 1
θ = π/4
__________

= -ln|cosθ| - 2θ/3 | [Evaluated from θ arctan(2/3) to π/4]

= [-ln|cos(π/4)| - (2/3)(π/4)]
- [-ln|cos(arctan(2/3))| - (2/3)(arctan(2/3)]

= [-ln(1/√2) - π/6] + [ln|cos(arccos(3/√13))| + (2/3)(arctan(2/3)]

= ln(√2) - π/6 + ln(3/√13) + (2/3)(arctan(2/3)

Answer 2

x dx/[x^2 + 4x + 13]

you write the denominator as

x^2 + 4x + 4 + 9 = (x+2)^2 + 9

let x+2 = 3 tan u: x = 3 tan u - 2

dx = 3 sec^2 u du

∫x dx/x^2 + 4x + 13

= ∫(3tan u - 2) du/ (9 tan^u + 9)

=> ∫(3 tan u - 2 )( 3 sec^u) du/ 9( sec^2 u)

=>∫ tanu du - 2/3∫ du

=> - ln cos u - 2/3 u

substituting u in terms of x

=> - ln [3/sqrt(x^2 + 4x + 13) - 2/3 arctan(x+2)/3 + c