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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

= 2 \cdot \int \frac{cos ( u )}{sin ( u ) + cos ( u )} d u

Rewrite using trig identities

= 2 \cdot \int \frac{1}{2} + \frac{cos ( u ) - sin ( u )}{2 ( sin ( u ) + cos ( 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

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

\int \frac{1}{2} d u = \frac{1}{2} u

\int \frac{cos ( u ) - sin ( u )}{2 ( sin ( u ) + cos ( u ) )} d u = \frac{1}{2} ln ∣ sin (u) + cos (u) ∣

= 2 (\frac{1}{2} u + \frac{1}{2} ln ∣ sin (u) + cos (u) ∣)

Substitute back

u = arcsin (x)

= 2 (\frac{1}{2} arcsin (x) + \frac{1}{2} ln ∣ sin (arcsin (x)) + cos (arcsin (x)) ∣)

Simplify

2 (\frac{1}{2} arcsin (x) + \frac{1}{2} ln ∣ sin (arcsin (x)) + cos (arcsin (x)) ∣) : 2 (\frac{1}{2} arcsin (x) + \frac{1}{2} ln x + 1 - x^{2})

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

Add a constant to the solution

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

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