Which the integral of 1/(7+8x^2)^(1/2)dx

Recall that

sec(t)^2 - tan(t)^2 = 1
sec(t)^2 = 1 + tan(t)^2

Let x = sqrt(7/8) * tan(t)
dx would be sqrt(7/8) * sec(t)^2 * dt

Now we have:

sqrt(7/8) * sec(t)^2 * dt / (7 + 8 * (7/8) * tan(t)^2)^(1/2) =>
sqrt(7/8) * sec(t)^2 * dt / (7 + 7 * tan(t)^2)^(1/2) =>
sqrt(7/8) * sec(t)^2 * dt / (sqrt(7) * sqrt(1 + tan(t)^2)) =>
sqrt(7) * sec(t)^2 * dt / (sqrt(7) * sqrt(8) * sqrt(sec(t)^2)) =>
sec(t)^2 * dt / (sqrt(8) * sec(t)) =>
sec(t) * dt / (2 * sqrt(2))

The integral of sec(t) * dt is ln|sec(t) + tan(t)| + C

(sqrt(2) / 4) * ln|sec(t) + tan(t)| + C

x = sqrt(7/8) * tan(t)
x / sqrt(7/8) = tan(t)
tan(t) = sqrt(8/7) * x

sec(t) = sqrt(1 + x^2 / (7/8)) = sqrt(1 + 8x^2 / 7) = sqrt((7 + 8x^2) / 7)

(sqrt(2) / 4) * ln|sqrt((7 + 8x^2) / 7) + sqrt(8/7) * x| + C =>
(sqrt(2) / 4) * ln|sqrt(7 + 8x^2) / sqrt(7) + sqrt(8) * x / sqrt(7)| + C =>
(sqrt(2) / 4) * (ln|sqrt(7 + 8x^2) + sqrt(8) * x| - ln|sqrt(7)|) + C

∫dx/√(7 + 8x²)

1/√8 * ∫dx/√(7/8 + x²)

Make the substitution x = √(7/8) * tan(t) where t = theta.

1/√8 * ∫sec(t) dt

1/√8 * ln|sec(t) + tan(t)| + C

1/√8 * ln|√(7 + 8x²) + x| + C

16x-3