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Topic starter
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.
Posted : 06/05/2010 8:24 pm
Topic starter
Sorted.
Posted : 06/05/2010 9:22 pm
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.
Posted : 06/05/2010 9:49 pm
42
Posted : 06/05/2010 9:50 pm
FAIL
Posted : 06/05/2010 9:52 pm
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
Posted : 06/05/2010 10:40 pm