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
    1 decade ago
    Favorite Answer

    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

  • 1 decade ago

    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
    1 decade ago

    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
    1 decade ago

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

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  • Anonymous
    4 years ago

    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)

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