Air resistance acting on a falling body can be taken into account by the approximate relation for the acceleration:

a=dv/dt=g-kv where k is constnat

Derive a formula for the velocity of the body as a function of time assuming it starts from rest (v=0 at t=0)

v=?

Integral of dv/(g - kv) = integral of dt

Integrate both sides, from time from 0 to t; and v from 0 to v, for an equation for t in terms of v. Then invert the equation for v(t)

It would be great to see how this is done. The prof's way of doing it is way confusing and intricate and I just started learning integrals.

This is why you'd need to do exercises in integration.

To give you a hint,
∫dv/(g-kv) = -log(g-kv)/k
This could be inferred from standard integrals:
∫dx/(a+bx) = (1/b)log(a+bx)

I agree. I suspect the Prof's method is confusing because you don't understand it....you learn by experience spaced over time. There is no substitute in calculus for hump, grunt, and strain...that is, practice, trial and error.

To derive a formula for the velocity of the body as a function of time, we can solve the differential equation

dv/dt = g - kv

Given that the body starts from rest (v = 0) at t = 0, we need to integrate both sides of the equation to find v as a function of t.

Let's start by rearranging the equation:

dv = (g - kv) dt

Next, we can separate the variables by moving the v terms to the left side and the t terms to the right side:

dv/(g - kv) = dt

To integrate, we need to rewrite the left side of the equation. We can use a technique called partial fraction decomposition to break it down:

1/(g - kv) = A/(g - kv)

Where A is a constant that we need to determine.

Let's multiply both sides of the equation by (g - kv) to get rid of the denominator:

1 = A

Now, substituting this value back into the equation:

dv/(g - kv) = dt

dv = (g - kv) dt

Now we can integrate both sides:

∫dv = ∫(g - kv) dt

v = ∫ (g - kv) dt

Integrating the right side of the equation:

v = ∫ g dt - ∫ kvt dt

The integral of a constant (g) with respect to time gives gt. And the integral of kvt with respect to time gives (1/2)kvt^2.

v = gt - (1/2)kvt^2 + C

Where C is a constant determined by the initial condition v = 0 at t = 0.

Since the body starts from rest (v = 0) at t = 0, we can substitute these initial conditions into the equation to solve for C:

0 = g(0) - (1/2)k(0)(0)^2 + C

0 = C

Therefore, the constant C is 0.

So the final formula for the velocity of the body as a function of time is:

v = gt - (1/2)kvt^2