What is the integral of (2x+6)/(sqrt(16-6x-x^2)) ?

The integral of (2x+6)/(sqrt(16-6x-x^2)) is -2sqrt(-x^2-6x+16)+C

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integral of (2x+6)/(sqrt(16-6x-x^2))

\int \frac{2 x + 6}{16 - 6 x - x ^{2}} d x

- 2 - x^{2} - 6 x + 16 + C

Solution steps

\int \frac{2 x + 6}{16 - 6 x - x ^{2}} d x

Expand

\frac{2 x + 6}{16 - 6 x - x ^{2}} : \frac{2 x}{16 - 6 x - x ^{2}} + \frac{6}{16 - 6 x - x ^{2}}

= \int \frac{2 x}{16 - 6 x - x ^{2}} + \frac{6}{16 - 6 x - 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}{16 - 6 x - x ^{2}} d x + \int \frac{6}{16 - 6 x - x ^{2}} d x

\int \frac{2 x}{16 - 6 x - x ^{2}} d x = 2 (- - x^{2} - 6 x + 16 - 3 arcsin (\frac{1}{5} (x + 3)))

\int \frac{6}{16 - 6 x - x ^{2}} d x = 6 arcsin (\frac{1}{5} (x + 3))

= 2 (- - x^{2} - 6 x + 16 - 3 arcsin (\frac{1}{5} (x + 3))) + 6 arcsin (\frac{1}{5} (x + 3))

Simplify

2 (- - x^{2} - 6 x + 16 - 3 arcsin (\frac{1}{5} (x + 3))) + 6 arcsin (\frac{1}{5} (x + 3)) : - 2 - x^{2} - 6 x + 16

= - 2 - x^{2} - 6 x + 16

Add a constant to the solution

= - 2 - x^{2} - 6 x + 16 + C

What is the integral of (2x+6)/(sqrt(16-6x-x^2)) ?
The integral of (2x+6)/(sqrt(16-6x-x^2)) is -2sqrt(-x^2-6x+16)+C