INTEGRATION arctanx ?

1 Answer
Steve M
May 30, 2018

# int \ arctan x \ dx = xarctan x - 1/2 ln(x^2+1) + C#

Explanation:

We seek:

# I = int \ arctan x \ dx #

We can apply Integration By Parts

Let # { (u,=arctanx, => (du)/dx,=1/(1+x^2)), ((dv)/dx,=1, => v,=x ) :}#

Then plugging into the IBP formula:

# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #

gives us

# int \ (arctanx)(1) \ dx = (arctanx)(x) - int \ (x)(1/(x^2+1)) \ dx #

# :. int \ arctan x \ dx = xarctan x - 1/2 \ int \ (2x)/(x^2+1) \ dx #
# " " = xarctan x - 1/2 ln|x^2+1| + C #

Noting that #x^2+1 gt 0 AA x in RR# we can write:

And so:

# int \ arctan x \ dx = xarctan x - 1/2 ln(x^2+1) + C #

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