OK, probably being a total numpty.

Trying to integrate;
1/(4(x^2)+1)
I keep getting
tan^-1(2x)+c

But the book says the answer is
(1/2)tan^-1(2x)+c

Can anyone show me where that 1/2 has come from?

Ta, Duane.

Sorted.

tan…du/(x^2+1)
So initially it becomes (1/4x)(1/(tan(x^2+1))…
4/(-2x^2)=2/x = 2/4 = 1/2
Hence the 1/2 tan^1(x^2+1)…1/2 tan^1(2x)…(& adding the +c).

Or something like that.
Been a few years since I did this stuff.
Uuuuurgh.

42

FAIL

where'd the tan come from?

So it starts with: 1/(4(x^2)+1)
Then i would then do: 1/((4x^2)+1)
Then: 1/((4/3x^3)+x)+c
Then: ((4/3x^3)+x)^-1)+c