Find the integral?

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