Asked by Jake
why does substituting u for (x^2)-1 work less neat when integrating (((x^2)-1)/x) dx then when integrating (x/((x^2)-1)) dx? How do you integrate the first one simply?
Answers
Answered by
Jake
Is it best to move the x up first or can you not do that?
Answered by
oobleck
if u = x^2-1, then du = 2x dx
That makes it perfect for integrating x/(x^2-1), since you then have
1/2 du/u
But, things aren't so good when you have (x^2-1)/x since that would give you 2u/du. Having du in the bottom is not the way things are done.
Besides, (x^2-1)/x is just x - 1/x
so you don't even need a substitution.
That makes it perfect for integrating x/(x^2-1), since you then have
1/2 du/u
But, things aren't so good when you have (x^2-1)/x since that would give you 2u/du. Having du in the bottom is not the way things are done.
Besides, (x^2-1)/x is just x - 1/x
so you don't even need a substitution.
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