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

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

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integral of 1/(1+sqrt(1-z^2))

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

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

Solution steps

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

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

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

Expand

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

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

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

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

Simplify

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

Add a constant to the solution

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

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