What is the antiderivative of tan^-1 (x)?

Question

^ means to the power of/exponent

So what I'm asking for is the antiderivative of inverse tangent, or arc tangent of x

Oh yeah, I know that the ^ doesn't apply here, but I meant notation-wise. Sorry for being vague.

Answer 1

In your case, ^ would NOT mean "to the power of", because "arctan" would mean functional inverse, not multiplicative.

Either way, to solve

Integral (tan^(-1)(x) dx)

You have to do this by parts.

Let u = tan^(-1)(x). dv = dx
du = 1/(1 + x^2) dx. v = x

Note that when doing integration by parts, you go

uv - Integral (v du)

Therefore,

xtan^(-1)(x) - Integral (x / (1 + x^2)) dx

To solve this new integration, substitution of u = 1 + x^2 will reveal that its integral is (1/2) ln |1 + x^2|, making our answer

xtan^(-1)(x) - (1/2) ln |1 + x^2| + C

As a reminder, tan^(-1)(x) is NOT the same as 1/(tan(x))

Answer 2

Int Arctg x dx. Integrating by parts =x*Arctgx-Int x/(1+x^2) =

= x*Arctgx -1/2 ln I 1+x^2I

Answer 3

puggy is right but no need for modulus 0f 1+x^2 since it is always positive