can you integrate the sum of the last two integral expressions?
u^2 +1 du
it started as the integral of...
and i completed the square to get the integral of...
Then put u = x-1, so that x = u+1, and thats how i got to where i am now... i always get stuck at this part...thanks
its like my brain freezes when i get an integral as such.
The integral of...
(3) [ (u du) / (u^2 +1) ]
you said it was the natural log, but if you do the natural log, dont you need a constant as the numerator?
the way i did it ended up using r = u^2+1 and the integral turns to...
(3) [ (u dr) / r (2u) ]
then the "u" cancels to get the integral of...
(3/2) [ dr / r ]
3/2 ln abs(r)
3/2 ln abs(u^2 +1)
3/2 ln abs( (x+1)^2 + 1)
3/2 ln abs(x^2 -2x +1)....which the function in the ln was the denominator all along...to me this seems wrong but the math seems correct...i probably missed something...(besides the constant at the end of course)