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

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

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

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

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

Solution steps

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

Simplify

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

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

Expand

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

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

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

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

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

Add a constant to the solution

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

\int 4 y^{2} d y

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 36})

\int \frac{1}{x ^{2} - 5} d x

n = 1 \sum \infty 5

\frac{\partial}{\partial y} (x^{3} yz e^{z})

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