What is the integral of 1/(1+(1-x^2)^{1/2)} ?

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

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integral of 1/(1+(1-x^2)^{1/2)}

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{cos ( u )}{1 + cos ( u )} d u

= \int \frac{cos ( u )}{1 + cos ( u )} d u

Multiply by the conjugate of

1 + cos (u) : \frac{sin ^{2} ( u )}{1 - cos ( u )}

= \int \frac{cos ( u )}{\frac{s i n ^{2} ( u )}{1 - c o s ( u )}} d u

Expand

\frac{cos ( u )}{\frac{s i n ^{2} ( u )}{1 - c o s ( u )}} : \frac{cos ( u )}{sin ^{2} ( u )} - \frac{cos ^{2} ( u )}{sin ^{2} ( u )}

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

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{cos ( u )}{sin ^{2} ( u )} d u - \int \frac{cos ^{2} ( u )}{sin ^{2} ( u )} d u

\int \frac{cos ( u )}{sin ^{2} ( u )} d u = - csc (u)

\int \frac{cos ^{2} ( u )}{sin ^{2} ( u )} d u = - cot (u) - u

= - csc (u) - (- cot (u) - u)

Substitute back

u = arcsin (x)

= - csc (arcsin (x)) - (- cot (arcsin (x)) - arcsin (x))

Simplify

- csc (arcsin (x)) - (- cot (arcsin (x)) - arcsin (x)) : - \frac{1}{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

\int y 9 - y^{2} d y

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\int \frac{3}{9 x ^{2} + 4} d x

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