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

Question

Cameron Alexander · Accepted Answer

# 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 #

Charles Arocha · Answer

undefined To find the integral of $ \frac{1}{(1+x^2)\sqrt{1-\arctan x}} $ with respect to $ x $, you can use substitution method. Let $ u = \arctan x $. Then $ du = \frac{1}{1+x^2}dx $. Substitute $ u $ and $ du $ into the integral:

$$
\int \frac{1}{(1+x^2)\sqrt{1-\arctan x}} dx = \int \frac{1}{\sqrt{1-u}} du
$$

This integral can be evaluated using a trigonometric substitution or by recognizing it as a standard integral. Let $ v = \sqrt{1-u} $, then $ u = 1 - v^2 $ and $ du = -2v \, dv $. Substitute $ u $ and $ du $ into the integral:

$$
\int \frac{1}{\sqrt{1-u}} du = \int \frac{-2v}{v} dv = -2\int dv = -2v + C
$$

Substitute $ v = \sqrt{1-u} $ back in terms of $ x $ and $ u $:

$$
= -2\sqrt{1-\arctan x} + C
$$

Therefore, the integral of $ \frac{1}{(1+x^2)\sqrt{1-\arctan x}} $ with respect to $ x $ is $ -2\sqrt{1-\arctan x} + C $, where $ C $ is the constant of integration.