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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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