Find #int 1/((1+x^2)sqrt(1-arctanx)) dx #?

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Answer 1 · 2017-12-13T16:28:40

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

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 #