I was wondering how I integrate this basic equation:

y' = 1 / (3 + 5x + x^2)

Any help would be great, thanks.

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- Apr 12th 2009, 04:13 AMPhatmatHow do I integrate this basic equation?
I was wondering how I integrate this basic equation:

y' = 1 / (3 + 5x + x^2)

Any help would be great, thanks. - Apr 12th 2009, 04:42 AMstapel
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.... (Wink) - Apr 12th 2009, 05:10 AMPhatmat
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.