Find #int 1/((1+x^2)sqrt(1-arctanx)) dx #?
1 Answer
Dec 13, 2017
# int \ 1/((1+x^2)sqrt(1-arctanx)) \ dx = - 2sqrt(1-arctanx) + C #
Explanation:
We seek:
# I = int \ 1/((1+x^2)sqrt(1-arctanx)) \ dx #
We can perform a substitution, Let:
# u = 1-arctanx => (du)/dx = -1/(x^2+1) #
Substituting into the integral we get:
# I = int \ (-1)/sqrt(u) \ du #
# \ \ = int \ (-1)/sqrt(u) \ du #
# \ \ = - \ int \ (1)/sqrt(u) \ du #
This is now a trivial integration, so sing the power rule:
# I = - 2sqrt(u) + C #
Finally, restoring the substitution:
# I = - 2sqrt(1-arctanx) + C #