Verified answer

dx / (x^5 * sqrt(16x^2 - 1))

x = (1/4) * sec(t)

dx = (1/4) * sec(t) * tan(t) * dt

(1/4) * sec(t) * tan(t) * dt / ((1/4)^5 * sec(t)^5 * sqrt(16 * (1/16) * sec(t)^2 - 1)) =>

(1/4) * sec(t) * tan(t) * dt / ((1/4)^5 * sec(t)^5 * sqrt(tan(t)^2)) =>

(1/4) * sec(t) * tan(t) * dt / ((1/4)^5 * sec(t)^5 * tan(t)) =>

dt / ((1/4)^4 * sec(t)^4) =>

256 * cos(t)^4 * dt

That's a little cleaner, isn't it?

256 * cos(2t/2)^4 * dt

256 * (cos(2t/2)^2)^2 * dt

256 * ((1/2) * (1 + cos(2t)))^2 * dt

256 * (1/4) * (1 + 2 * cos(2t) + cos(2t)^2) * dt =>

64 * (1 + 2 * cos(2t) + cos(2t)^2) * dt =>

64 * dt + 128 * cos(2t) * dt + 64 * cos(2t)^2 * dt =>

64 * dt + 128 * cos(2t) * dt + 64 * cos(4t/2)^2 * dt =>

64 * dt + 128 * cos(2t) * dt + 64 * (1/2) * (1 + cos(4t)) * dt =>

64 * dt + 128 * cos(2t) * dt + 32 * (1 + cos(4t)) * dt =>

64 * dt + 128 * cos(2t) * dt + 32 * dt + 32 * cos(4t) * dt =>

96 * dt + 128 * cos(2t) * dt + 32 * cos(4t) * dt

Now we can integrate

96t + 64 * sin(2t) + 8 * sin(4t) + C

Now, we have 2 options. We can either back-substitute in order to have our terms in the form of x, or we can change our bounds to a form of t (that's what I'm going to do, simply because it would be easier)

x = (1/4) * sec(t)

x = sqrt(2)/4

sqrt(2)/4 = (1/4) * sec(t)

sqrt(2) = sec(t)

pi/4 = t

x = 1/2

1/2 = (1/4) * sec(t)

2 = sec(t)

t = pi/3

Integrate from pi/4 to pi/3

96t + 64 * sin(2t) + 8 * sin(4t) + C

96 * (pi/3 - pi/4) + 64 * (sin(2 * pi/3) - sin(2 * pi/4)) + 8 * (sin(4 * pi/3) - sin(4 * pi/4)) =>

96 * (4pi/12 - 3pi/12) + 64 * (sin(2pi/3) - sin(pi/2)) + 8 * (sin(4pi/3) - sin(pi)) =>

96 * (pi/12) + 64 * (sqrt(3)/2 - 1) + 8 * (-sqrt(3)/2 - 0) =>

8 * pi + 32 * sqrt(3) - 64 - 4 * sqrt(3) =>

8 * pi + 28 * sqrt(3) - 64

There you go.

Now you apply the trig relationships of:

sec^2 theta = tan^2 theta + 1

derivative of tan theta = sec^2 theta

Substitute and finish

You're doing good so far. Keep going.

(4⁵/4) ∫ sec(u)tan(u) du / (sec⁵(u)√(sec²(u) - 1))

4⁴ ∫ sec(u) tan(u) du / (sec⁵(u)tan(u))

4⁴ ∫ cos⁴(u) du

4⁴ ∫ (cos²(u))² du

4⁴ ∫ ((1 + cos(2u)) / 2)² du

Expand that, use the power reduction again, some more substitutions to deal with the constants, and you'll have elementary integrals.

I'd highly recommend changing the limits to u (or theta) rather than trying to back-substitute.

including to Moja19's answer: Partial fractions on "(2x-3)/(x-a million)^2 = A/(x-a million) + B/(x-a million)^2" may be accomplished slightly greater truthfully via taking cut back as x->a million and noting left is going to -a million/(x-a million)^2 and top is going to B/(x-a million)^2, so B=-a million. A is discovered via placing x=0 and fixing. This works super, whilst there are not any repeated roots, yet helps in each and every case. right here we had a million equation and a million unknown instead of two equations 2 unknowns.

∫x⁵√(16x² - 1) dx

x = (1/4)secθ

dx = (1/4)secθtanθ dθ

(1/4)⁶ * ∫(sec⁵θ * tanθ) * (secθtanθ) dθ

(1/4)⁶ * ∫(sec⁶θ * tan²θ) dθ

I'll let you take it from here.