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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (5 x (sin (x) + cos (x)))

7 x^{2} y + y^{^{'}} = 6 x^{2}

y^{^{'''}} - 5 y^{^{''}} - 6 y^{^{'}} = 0

\frac{d}{d t} (t)

x y^{^{'}} - 4 y = x

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