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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \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

= \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) ∣

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

Substitute back

u = arcsin (x)

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

Simplify

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

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

Add a constant to the solution

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

\int 3 x (x^{2} + 8)^{2} d x

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\int \frac{1}{x ( x ^{6} + 1 )} d x

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

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