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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u - 2}{4 - u ^{2}} d u

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

Expand

\frac{u - 2}{4 - u ^{2}} : \frac{u}{4 - u ^{2}} - \frac{2}{4 - u ^{2}}

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

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

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

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

Substitute back

u = x + 2

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

Simplify

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

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

Add a constant to the solution

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

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