Volumes of Revolution Problem ?? Help

Ok the problem is :

Find the volume of the solid obtained by rotating the region under the graph of the function
f(x) = 1/(x+1) about the x-axis over the interval [0,2]

the answer i came up with is ( ( - pi/3 ) - ( - pi ) )

is this correct

Using the disk method,
V = π ∫₀² (1/(x + 1))² dx

Letting u = x + 1,
du = dx
And the new limits are
upper: u = 2 + 1 = 3
lower: u = 0 + 1 = 1

V = π ∫₁³ 1/u² du
V = π ∫₁³ u⁻² du
V = π [-u⁻¹]₁³
V = π [-1/u]₁³
V = π [-1/3 - (-1)]
V = π [2/3] = 2π/3

Your answer was
( (-π/3) - (-π) )
(-π/3 + π)
π (-1/3 + 1)
π (2π/3)
2π/3

So yes, you are correct.