What is the integral of (x+3)sqrt(4-x^2) ?

The integral of (x+3)sqrt(4-x^2) is -1/3 (4-x^2)^{3/2}+6(arcsin(1/2 x)+1/2 sin(2arcsin(1/2 x)))+C

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integral of (x+3)sqrt(4-x^2)

\int (x + 3) 4 - x^{2} d x

- \frac{1}{3} (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x))) + C

Solution steps

\int (x + 3) 4 - x^{2} d x

Expand

(x + 3) 4 - x^{2} : x 4 - x^{2} + 3 4 - x^{2}

= \int x 4 - x^{2} + 3 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 x 4 - x^{2} d x + \int 3 4 - x^{2} d x

\int x 4 - x^{2} d x = - \frac{1}{3} (4 - x^{2})^{\frac{3}{2}}

\int 3 4 - x^{2} d x = 6 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x)))

= - \frac{1}{3} (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x)))

Add a constant to the solution

= - \frac{1}{3} (4 - x^{2})^{\frac{3}{2}} + 6 (arcsin (\frac{1}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} x))) + C

ln (15)

x \to 9 lim (ln (x - 9))

\frac{d}{d x} (\frac{( x + 1 )}{( x + 2 )})

x^{2}, 4, [0, 2]

\int_{0}^{4} \frac{1}{16 + t ^{2}} d t

What is the integral of (x+3)sqrt(4-x^2) ?
The integral of (x+3)sqrt(4-x^2) is -1/3 (4-x^2)^{3/2}+6(arcsin(1/2 x)+1/2 sin(2arcsin(1/2 x)))+C