What is the integral of 1/((2-x)sqrt(1-x^2)) ?

The integral of 1/((2-x)sqrt(1-x^2)) is 2/(sqrt(3))arctan(1/(sqrt(3))(2tan((arcsin(x))/2)-1))+C

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integral of 1/((2-x)sqrt(1-x^2))

\int \frac{1}{( 2 - x ) 1 - x ^{2}} d x

\frac{2}{3} arctan (\frac{1}{3} (2 tan (\frac{arcsin ( x )}{2}) - 1)) + C

Solution steps

\int \frac{1}{( 2 - x ) 1 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{2 - sin ( u )} d u

= \int \frac{1}{2 - sin ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2} + 1 - v} d v

= \int \frac{1}{v ^{2} + 1 - v} d v

Complete the square

v^{2} + 1 - v : (v - \frac{1}{2})^{2} + \frac{3}{4}

= \int \frac{1}{( v - \frac{1}{2} ) ^{2} + \frac{3}{4}} d v

Apply u-substitution

: \int \frac{4}{4 w ^{2} + 3} d w

= \int \frac{4}{4 w ^{2} + 3} d w

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{1}{4 w ^{2} + 3} d w

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 3 ( t ^{2} + 1 )} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{2 3} \cdot \int \frac{1}{t ^{2} + 1} d t

Use the common integral:

\int \frac{1}{t ^{2} + 1} d t = arctan (t)

= 4 \cdot \frac{1}{2 3} arctan (t)

Substitute back

= 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (tan (\frac{arcsin ( x )}{2}) - \frac{1}{2}))

Simplify

4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (tan (\frac{arcsin ( x )}{2}) - \frac{1}{2})) : \frac{2}{3} arctan (\frac{1}{3} (2 tan (\frac{arcsin ( x )}{2}) - 1))

= \frac{2}{3} arctan (\frac{1}{3} (2 tan (\frac{arcsin ( x )}{2}) - 1))

Add a constant to the solution

= \frac{2}{3} arctan (\frac{1}{3} (2 tan (\frac{arcsin ( x )}{2}) - 1)) + C

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What is the integral of 1/((2-x)sqrt(1-x^2)) ?
The integral of 1/((2-x)sqrt(1-x^2)) is 2/(sqrt(3))arctan(1/(sqrt(3))(2tan((arcsin(x))/2)-1))+C