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## Homework Statement

Find [tex]\int[/tex](x+2)dx/sqrt(3+2x-x

^{2})

## Homework Equations

[tex]\int[/tex]du/sqrt(a

^{2}- u

^{2}) = sin

^{-1}u/a

## The Attempt at a Solution

I began by completing the square:

3+2x-x

^{2}= 4 -(x

^{2}-2x+1)

So, 4-(x-1)

^{2}= a

^{2}-u

^{2}and a=2 and u=(x-1)

Further, since x=(u+1), dx=du and (x+2)=(u+3)

Substituting, I got:

[tex]\int[/tex](x+2)dx/sqrt(3+2x-x

^{2}) = [tex]\int[/tex](u+3)du/sqrt(a

^{2}-u

^{2})

= [tex]\int[/tex]udu/sqrt(a

^{2}-u

^{2}) + 3[tex]\int[/tex]du/sqrt(a

^{2}-u

^{2})

This is where I get confused. I know that 3[tex]\int[/tex]du/sqrt(a

^{2}-u

^{2}) = 3sin

^{-1}u/a

But, according to my book, [tex]\int[/tex]udu/sqrt(a

^{2}-u

^{2}) = -sqrt(a

^{2}-u

^{2}) and I don't understand why.

Can someone offer an explanation? Thanks.