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

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

We seek:

We can perform a substitution, Let:

Substituting into the integral we get:

This is now a trivial integration, so sing the power rule:

Finally, restoring the substitution:

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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