INTEGRATION arctanx ?
1 Answer
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
And so:
# int \ arctan x \ dx = xarctan x - 1/2 ln(x^2+1) + C #