The integral of (5x+8) / (x^2+3x+2) from 0 to 1

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Mathematics

Cuddle Pig

23 June, 17:23

The integral of (5x+8) / (x^2+3x+2) from 0 to 1

+1

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Luca Benton · Answer 1

Compute the definite integral:

integral_0^1 (5 x + 8) / (x^2 + 3 x + 2) dx

Rewrite the integrand (5 x + 8) / (x^2 + 3 x + 2) as (5 (2 x + 3)) / (2 (x^2 + 3 x + 2)) + 1 / (2 (x^2 + 3 x + 2)):

= integral_0^1 ((5 (2 x + 3)) / (2 (x^2 + 3 x + 2)) + 1 / (2 (x^2 + 3 x + 2))) dx

Integrate the sum term by term and factor out constants:

= 5/2 integral_0^1 (2 x + 3) / (x^2 + 3 x + 2) dx + 1/2 integral_0^1 1 / (x^2 + 3 x + 2) dx

For the integrand (2 x + 3) / (x^2 + 3 x + 2), substitute u = x^2 + 3 x + 2 and du = (2 x + 3) dx.

This gives a new lower bound u = 2 + 3 0 + 0^2 = 2 and upper bound u = 2 + 3 1 + 1^2 = 6: = 5/2 integral_2^6 1/u du + 1/2 integral_0^1 1 / (x^2 + 3 x + 2) dx

Apply the fundamental theorem of calculus.

The antiderivative of 1/u is log (u) : = (5 log (u)) / 2 right bracketing bar _2^6 + 1/2 integral_0^1 1 / (x^2 + 3 x + 2) dx

Evaluate the antiderivative at the limits and subtract.

(5 log (u)) / 2 right bracketing bar _2^6 = (5 log (6)) / 2 - (5 log (2)) / 2 = (5 log (3)) / 2: = (5 log (3)) / 2 + 1/2 integral_0^1 1 / (x^2 + 3 x + 2) dx

For the integrand 1 / (x^2 + 3 x + 2), complete the square:

= (5 log (3)) / 2 + 1/2 integral_0^1 1 / ((x + 3/2) ^2 - 1/4) dx

For the integrand 1 / ((x + 3/2) ^2 - 1/4), substitute s = x + 3/2 and ds = dx.

This gives a new lower bound s = 3/2 + 0 = 3/2 and upper bound s = 3/2 + 1 = 5/2: = (5 log (3)) / 2 + 1/2 integral_ (3/2) ^ (5/2) 1 / (s^2 - 1/4) ds

Factor - 1/4 from the denominator:

= (5 log (3)) / 2 + 1/2 integral_ (3/2) ^ (5/2) 4 / (4 s^2 - 1) ds

Factor out constants:

= (5 log (3)) / 2 + 2 integral_ (3/2) ^ (5/2) 1 / (4 s^2 - 1) ds

Factor - 1 from the denominator:

= (5 log (3)) / 2 - 2 integral_ (3/2) ^ (5/2) 1 / (1 - 4 s^2) ds

For the integrand 1 / (1 - 4 s^2), substitute p = 2 s and dp = 2 ds.

This gives a new lower bound p = (2 3) / 2 = 3 and upper bound p = (2 5) / 2 = 5:

= (5 log (3)) / 2 - integral_3^5 1 / (1 - p^2) dp

Apply the fundamental theorem of calculus.

The antiderivative of 1 / (1 - p^2) is tanh^ (-1) (p):

= (5 log (3)) / 2 + (-tanh^ (-1) (p)) right bracketing bar _3^5

Evaluate the antiderivative at the limits and subtract. (-tanh^ (-1) (p)) right bracketing bar _3^5 = (-tanh^ (-1) (5)) - (-tanh^ (-1) (3)) = tanh^ (-1) (3) - tanh^ (-1) (5):

= (5 log (3)) / 2 + tanh^ (-1) (3) - tanh^ (-1) (5)

Which is equal to:

Answer: = log (18)