- #1
jesuslovesu
- 198
- 0
Alright so, I'm trying to take the integral of 3/2 / (u^2 + 3/4)
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]
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]