I was wondering how I integrate this basic equation:
y' = 1 / (3 + 5x + x^2)
Any help would be great, thanks.
Complete the square in the denominator to convert to one of the inverse hyperbolic trig forms, which I believe works out as follows:
. . . . .$\displaystyle \left(x\, +\, \frac{5}{2}\right)^2\, -\, \left(\frac{\sqrt{13}}{2}\right)^2$
Then see if you can find a way to use the derivative formula you memorized for the inverse hyperbolic tangent....
This was actually for an eingineering problem. The 3,5,1 weren't the actual values, I had values that were very large and irregular:
e.g. 1/4.12x10^13 - (1.284x10^7)x + x^2
Is there an easier way to integrate this? I'm not too sure about what you described above, stapel :P.