Integrals: #1 Help with fraction #2 Moment of inertia

[togo]

Homework Statement

25-2-EX9
The time rate of change of the displacement (velocity) of a robot arm is ds/dt = 8t/(t^2 + 4)^2. Find the expression for the displacement as a function of time if s = -1 m when t = 0 s.
26-5-9
Find the moment of inertia of a plate covering the first-quadrant region bounded by y^2 = x, x = 9 and the x-axis with respect to x-axis.

Homework Equations

25-2-EX9 unknown
26-5-9 Ix =k \int_c^d y^2(x_2 - x_1)dy

The Attempt at a Solution

25-2-EX9
This is a book example, so the solution is here. I am hung up on one step though.
\int ds = \int \frac{8tdt}{(t^2 + 4)^2} = 4 \int (t^2 + 4)^{-2}(2t dt)

This next step throws me off. What happened to (2t dt)??

it just disappears. Can someone explain? Thanks.

s = 4(\frac{1}{-1})(t^2 + 4)^{-1} + C

26-5-9
Ix =k \int_c^d y^2(x_2 - x_1)dy

9ky^2dx = 9kx = 9k\frac{1}{2}x^2

4.5x^2 = 4.5 * 3^2

4.5*9 = 40.5However, the answer is \frac{162}{5}k

so where'd I go wrong? thanks.

 

[eumyang]

The Attempt at a Solution

25-2-EX9
This is a book example, so the solution is here. I am hung up on one step though.
\int ds = \int \frac{8tdt}{(t^2 + 4)^2} = 4 \int (t^2 + 4)^{-2}(2t dt)

This next step throws me off. What happened to (2t dt)??

it just disappears. Can someone explain? Thanks.

s = 4(\frac{1}{-1})(t^2 + 4)^{-1} + C

U-substitution was used without really showing it. Try letting u = t²+4 and rewrite the integral in terms of u and see what happens.

 

[togo]

which formula would that be?

 

[togo]

any suggestions?

 

[togo]

Problem with derivative

accidental post.