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

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

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

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

- \frac{1}{x} - \frac{1 - x ^{2}}{x} - arcsin (x) + C

Solution steps

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

\frac{1}{1 - 1 - x ^{2}} = \frac{1 + 1 - x ^{2}}{( 1 ) ^{2} - ( 1 - x ^{2} ) ^{2}}

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

Expand

\frac{1 + 1 - x ^{2}}{1 ^{2} - ( 1 - x ^{2} ) ^{2}} : \frac{1}{1 ^{2} - ( 1 - x ^{2} ) ^{2}} + \frac{1 - x ^{2}}{1 ^{2} - ( 1 - x ^{2} ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= - \frac{1}{x} - \frac{1 - x ^{2}}{x} - arcsin (x)

Add a constant to the solution

= - \frac{1}{x} - \frac{1 - x ^{2}}{x} - arcsin (x) + C

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