- #1

- 198

- 0

according to the book the answer is sqrt(3) * arctan(2u/sqrt(3))

but when I try to get the integral I get: 2*arctan(2u/sqrt(3)) -- I don't see any way to take a sqrt(3) out of the function.

The graphs look about the same except they are shifted vertically apart. Is this just a constant issue (are the answers basically equivilent)? Or did I mess up my algebra?

1

[tex]

\frac{3/2}{u^2+3/4}

[/tex]

2

[tex]3/2 * \frac{1}{(4u^2+3)/4}[/tex]

3

[tex]3/2*4 * \frac{1}{4u^2+3}[/tex]

4

[tex]6* \frac{1}{3(4/3u^2+1)}[/tex]