Find the integral?

Evaluate the integral 7/(w^2+3w+3) dw.

Here's my work:

I've completed the square and came with this:

integral of 7/[(w+3/2)^2+3/4] dw

where u=w+3/2 and du=dw.

After I substitute, I got

integral of 7/(u^2+3/4) du.

From here, I don't know what to do. I know that I'm supposed to do the inverse tangent, but I don't know how. Can anyone please tell me how to get to the correct answer from here? Please show your work. Thanks.

2 Answers

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  • 1 year ago
    Favourite answer

    Factor the 7 on top out of the integral and let a = sqrt(3/4) = sqrt(3)/2 and then you have the integral:

    ∫ 7/(u² + 3/4) du = 7 ∫ 1/(u² + a²) du

    That's a standard form, occurring often enough to memorize:

    ∫ 1/(x² + a²) dx = (1/a) tan⁻¹ (x/a) + C

    So, your integral becomes:

    = (7/a) tan⁻¹ (u/a) + C

    Since 1/a = 2/√3 and u = w + 3/2

    ∫ 1/(w² + 3w + 3) dw = (14 / √3) tan⁻¹ [(2w + 3) / √3] + C

    - - - - - Edit: Getting that 1/(x² + a²) integral from "scratch":

    Well, with "scratch" defined as knowing that (arctan x)' = 1/(x² + 1), anyway...

    Given that derivative, what you need is a substitution to get 1/(u² + a²) into 1/(v² + 1) form, times a constant that can be factored out. Divide top and bottom by a² to do that:

    1/(u² + a²) = (1/a²) / [(u/a)² + 1]

    Then let v = u/a, dv = (1/a) du, so that

    ∫ 1/(u² + a²) du = ∫ (1/a²) * 1/(v² + 1) * a dv

    (1/a) ∫ 1/(v² + 1) dv = (1/a) tan⁻¹ (v) + C

    = (1/a) tan⁻¹ (u/a) + C

  • Factor out 3/4 from the denominator:

    7 * du / ((3/4) * ((4/3) * u^2 + 1)) =>

    7 * (4/3) * du / ((4/3) * u^2 + 1)

    Now we can use a trig substitution

    1 + (4/3) * u^2 = 1 + tan(t)^2

    (4/3) * u^2 = tan(t)^2

    (4 * 3/9) * u^2 = tan(t)^2

    (2/3) * sqrt(3) * u = tan(t)

    (2/3) * sqrt(3) * du = sec(t)^2 * dt

    du = (3 / (2 * sqrt(3))) * sec(t)^2 * dt

    7 * (4/3) * du / ((4/3) * u^2 + 1) =>

    (28/3) * (sqrt(3)/2) * sec(t)^2 * dt / (tan(t)^2 + 1) =>

    (14/sqrt(3)) * sec(t)^2 * dt / sec(t)^2 =>

    (14/sqrt(3)) * dt

    Integrate

    (14/sqrt(3)) * t + C

    (2/3) * sqrt(3) * u = tan(t)

    (2 * sqrt(3) / 3) * u = tan(t)

    arctan((2 * sqrt(3) / 3) * u) = t

    (14/sqrt(3)) * arctan((2/sqrt(3)) * u) + C

    u = w + 3/2 = (1/2) * (2w + 3)

    (14/sqrt(3)) * arctan((2/sqrt(3)) * (1/2) * (2w + 3)) + C =>

    (14/sqrt(3)) * arctan((1/sqrt(3)) * (2w + 3)) + C

    Or with denominators rationalized

    (14sqrt(3)/3) * arctan((sqrt(3)/3) * (2w + 3)) + C

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