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

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

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

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

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

Solution steps

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

Expand

\frac{1 + x}{4 - x ^{2}} : \frac{1}{4 - x ^{2}} + \frac{x}{4 - x ^{2}}

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

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

\int \frac{1}{4 - x ^{2}} d x = arcsin (\frac{1}{2} x)

\int \frac{x}{4 - x ^{2}} d x = - 4 - x^{2}

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

Add a constant to the solution

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

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