ok came across a bump.
[integral] dx/ x^2 - 6x + 13
so i have to complete the square, so
x^2 - 6x + 13 = 0
x^2 - 6x = -13
wait before i continue I always thought anything that was negative and squared it become positive but now my calculator is telling me differently.
the...
oh ok and i just made up the example so if i have a number higher than 1 I need to divide that number out so ex. 5x^2 + 6x + 5 and then id divide the whole thing by 5.
I don't know what that means, since i work with the square root do I not pay attention to it until finally getting the final answer. ex.
x^3 + 6x^2 - 4
then
x^3 + 6x^2 = 4
x^3 + 6x^2 + 9 = 4 + 9
and so on
sorry if i was irritating you it has been a year since I've done any math and I am jumping into calc 2, but i finally got it and I have done most of the rest, just three left which I am working on now, so thank you for all your help and putting up with me :)
[integral] x cox^2 x dx
u= x
du = dx
dv= cos^2
v= (sin(2x)) / 4 + x/2
x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du
x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c
i know the answer is
x (sin(2x)) / 4 + cos(2x) / 8 + x^2 / 4 + c
but I got an extra x/2