### SOLUTIONS TO U-SUBSTITUTION

SOLUTION 8 : Integrate . Let

u = x2+4x-3

so that

du = (2x+4) dx = 2 (x+2) dx ,

or

(1/2) du = (x+2) dx .

Substitute into the original problem, replacing all forms of x, getting      .

SOLUTION 9 : Integrate . Let

u = x2+1

so that

du = 2x dx ,

or

(1/2) du = x dx .

Substitute into the original problem, replacing all forms of x, getting     (Recall that .)  .

SOLUTION 10 : Integrate . Let so that .

Substitute into the original problem, replacing all forms of x, getting    .

SOLUTION 11 : Integrate . Let so that (Don't forget to use the chain rule.) ,

or .

Substitute into the original problem, replacing all forms of x, getting  (Do not make the following VERY COMMON MISTAKE : . Why is this INCORRECT ?) .

Now make another substitution. Let

w = -u

so that

dw = (-1) du ,

or

(-1) dw = du .

Substitute into the problem, replacing all forms of u, getting     .

SOLUTION 12 : Integrate . Let

u = x2

so that

du = 2x dx ,

or

(1/2) du = x dx .

In addition, the range of x-values is ,

so that the range of u-values is ,

or .

Substitute into the original problem, replacing all forms of x and the x-values, getting      = (-1/2)( (-1) - (1) )

= (-1/2)( -2)

= 1 .

SOLUTION 13 : Integrate . Let

u = x-1

so that

du = (1) dx = dx .

In addition, we can "back substitute" with

x = u+1 .

Substitute into the original problem, replacing all forms of x, getting      .