Find $\int \frac{1}{(1 + x^{2}) \sqrt{1 - arctan x}} d x$ $int 1/((1+x^2)sqrt(1-arctanx)) dx$ ?

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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:

$I = - 2sqrt(1-arctanx) + C$

Find int 1/((1+x^2)sqrt(1-arctanx)) dx ∫1(1+x2)√1−arctanxdxint 1/((1+x^2)sqrt(1-arctanx)) dx ?