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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 2 (sin (u) + 1) d u

= \int 2 (sin (u) + 1) d u

Take the constant out:

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

= 2 \cdot \int sin (u) + 1 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 sin (u) d u + \int 1 d u)

\int sin (u) d u = - cos (u)

\int 1 d u = u

= 2 (- cos (u) + u)

Substitute back

u = arcsin (\frac{1}{2} x)

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

Simplify

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

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

Add a constant to the solution

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

e^{x}

\frac{\partial}{\partial x} (x^{3} + y z^{2})

x \to 3 lim (\frac{1}{ln ^{2} ( \frac{x}{3} )})

(x^{2} - 2)^{2}

\frac{\partial}{\partial z} (x y + yz)

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