What is the integral of (1+sqrt(9-x^2)) ?

The integral of (1+sqrt(9-x^2)) is x+9/2 (arcsin(1/3 x)+1/2 sin(2arcsin(1/3 x)))+C

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integral of (1+sqrt(9-x^2))

\int (1 + 9 - x^{2}) d x

x + \frac{9}{2} (arcsin (\frac{1}{3} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} x))) + C

Solution steps

\int 1 + 9 - 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 1 d x + \int 9 - x^{2} d x

\int 1 d x = x

\int 9 - x^{2} d x = \frac{9}{2} (arcsin (\frac{1}{3} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} x)))

= x + \frac{9}{2} (arcsin (\frac{1}{3} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} x)))

Add a constant to the solution

= x + \frac{9}{2} (arcsin (\frac{1}{3} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} x))) + C

\frac{d}{d x} (4 x^{3} - 6 x)

3 - e^{2 t}

f (x) = 2 - x

f (x) = \frac{5 x + 1}{x - 1}

y = x^{2} - 4, y = 8 - 2 x^{2}

What is the integral of (1+sqrt(9-x^2)) ?
The integral of (1+sqrt(9-x^2)) is x+9/2 (arcsin(1/3 x)+1/2 sin(2arcsin(1/3 x)))+C