Need a little math help..! Antiderivate..?

Question

Hiw can I find a antidrdivate of this function..

1/(1+x^2)

Thanks..

Answer 1

let x = tan t
dx = sec^2 t dt

now 1+x^2 = sec^2 t

so int (1/(1+x^2) dx) = int(sec^2t dt/sec^2 t) = int dt

= t +C = tan^1 x + C

Answer 2

Standard integral: That's arctan x + C.
So, how do you know that the derivative of
arctan x is 1/(1+x²)?
Well, write y = arctan x.
Then x = tan y. Differentiate both sides to get.
1 = sec² y dy/dx.
dy/dx = 1/sec² y = 1/(1 + tan² y) = 1/(1+x²).

Answer 3

∫[1/(1+x²)]dx = arctan(x) + C

Answer 4

Look up trig integrals. I think its like the inverse tangent.