What is the integral of 1/(x+sqrt(4-x^2)) ?

The integral of 1/(x+sqrt(4-x^2)) is 1/2 arcsin(1/2 x)+1/2 ln((|x+sqrt(4-x^2)|)/2)+C

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integral of 1/(x+sqrt(4-x^2))

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

\frac{1}{2} arcsin (\frac{1}{2} x) + \frac{1}{2} ln (\frac{x + 4 - x ^{2}}{2}) + C

Solution steps

\int \frac{1}{x + 4 - 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 (\frac{1}{2} x)

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

Simplify

\frac{1}{2} arcsin (\frac{1}{2} x) + \frac{1}{2} ln sin (arcsin (\frac{1}{2} x)) + cos (arcsin (\frac{1}{2} x)) : \frac{1}{2} arcsin (\frac{1}{2} x) + \frac{1}{2} ln (\frac{x + 4 - x ^{2}}{2})

= \frac{1}{2} arcsin (\frac{1}{2} x) + \frac{1}{2} ln (\frac{x + 4 - x ^{2}}{2})

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{1}{2} x) + \frac{1}{2} ln (\frac{x + 4 - x ^{2}}{2}) + C

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What is the integral of 1/(x+sqrt(4-x^2)) ?
The integral of 1/(x+sqrt(4-x^2)) is 1/2 arcsin(1/2 x)+1/2 ln((|x+sqrt(4-x^2)|)/2)+C