What does ∫1/(1+x^2) equal?
I feel like it has something to do with u substitution and a natural log but I can't seem to figure this one out and it's been bugging me all day. If you could show the steps to the answer that would be great, thanks!
Update:Thanks, I thought it was tan^-1 but the back of the book had something else, it just really confused me.
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∫1/(1+x^2) dx = tan^-1(x) + c
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You can use u substitution to prove it.
x = tan u
dx = sec^2 u du
∫1/(1+x^2) = ∫sec^2 u/(1+tan^2 u) du = ∫du = u+c = tan^-1(x) + c