Integral of Inverse Trig Function?

Integration by parts, followed by U-substitution.

Since I know you're an expert I'll just not explain each step and do it.

∫(arctan(x) dx)

Let u = arctan(x). dv = dx

du = 1/[1 + x^2] dx. v = x

x arctan(x) - ∫ (x/[1 + x^2] dx )

x arctan(x) - ∫ (1/[1 + x^2] x dx )

Let u = 1 + x^2.

du = 2x dx

(1/2) du = x dx

x arctan(x) - ∫ ( (1/u) (1/2) du )

x arctan(x) - (1/2) ∫ ( (1/u) du )

x arctan(x) - (1/2)ln|u| + C

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

x arctan(x) - (1/2)ln(1 + x^2) + C

The complex exponential version of ArcTan(x) is:

(1/2) i (Log(1 - i x) - Log(1 + i x))

The integral of this is

(1/2) x i (Log(1 - i x) - Log(1 + i x)) - (1/2) Log(1 + x²)

But the first two terms is really x ArcTan(x), so that the answer is:

x ArcTan(x) - (1/2) Log(1 + x²)

Wikipedia has a section in their article called "Indefinite integrals of inverse trigonometric functions"

It has a proof using Integration By Parts method for arcsin. This would be identical to yours.

Hint: IBP with u = arctan x , dv = dx.....( x arctan x - [1/2] ln { x² +1} + C)

because of the fact you calculated it by using hand incorrectly. If I re-differentiate your consequence, I finally end up with: (2*x^2 - (x^2 + 25)*ln(x^2 + 25))/(2*x^4 + 50*x^2) If I re-differentiate the TI-89 calculator's consequence, I finally end up with: a million/(x^2+25)