Integral of Inverse Trig Function?

Calculate the integral ∫arctan(x) dx.

I know the answer because I can look it up in the table of integrals at the back of a calculus book, but I am drawing a blank trying to work it out.

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• Puggy
Lv 7

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)

• Zach
Lv 5

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.

• ted s
Lv 7