What is the integral of (2x+6)/(sqrt(9-x^2)) ?

The integral of (2x+6)/(sqrt(9-x^2)) is -2sqrt(9-x^2)+6arcsin(1/3 x)+C

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integral of (2x+6)/(sqrt(9-x^2))

\int \frac{2 x + 6}{9 - x ^{2}} d x

- 2 9 - x^{2} + 6 arcsin (\frac{1}{3} x) + C

Solution steps

\int \frac{2 x + 6}{9 - x ^{2}} d x

Expand

\frac{2 x + 6}{9 - x ^{2}} : \frac{2 x}{9 - x ^{2}} + \frac{6}{9 - x ^{2}}

= \int \frac{2 x}{9 - x ^{2}} + \frac{6}{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 \frac{2 x}{9 - x ^{2}} d x + \int \frac{6}{9 - x ^{2}} d x

\int \frac{2 x}{9 - x ^{2}} d x = - 2 9 - x^{2}

\int \frac{6}{9 - x ^{2}} d x = 6 arcsin (\frac{1}{3} x)

= - 2 9 - x^{2} + 6 arcsin (\frac{1}{3} x)

Add a constant to the solution

= - 2 9 - x^{2} + 6 arcsin (\frac{1}{3} x) + C

What is the integral of (2x+6)/(sqrt(9-x^2)) ?
The integral of (2x+6)/(sqrt(9-x^2)) is -2sqrt(9-x^2)+6arcsin(1/3 x)+C