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))

Simplify

= - \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

\int \frac{1}{( x ^{2} + 1 ) ^{3}} d x

\int sec^{2} (x) (1 + tan (x)) d x

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

\frac{1}{x ^{2}}

\int \frac{1 - x}{9 - 4 x ^{2}} d x

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