What is the integral of (3x+1)/(sqrt(-5-6x-x^2)) ?

The integral of (3x+1)/(sqrt(-5-6x-x^2)) is -3sqrt(-x^2-6x-5)-8arcsin(1/2 (x+3))+C

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integral of (3x+1)/(sqrt(-5-6x-x^2))

\int \frac{3 x + 1}{- 5 - 6 x - x ^{2}} d x

- 3 - x^{2} - 6 x - 5 - 8 arcsin (\frac{1}{2} (x + 3)) + C

Solution steps

\int \frac{3 x + 1}{- 5 - 6 x - x ^{2}} d x

Expand

\frac{3 x + 1}{- 5 - 6 x - x ^{2}} : \frac{3 x}{- 5 - 6 x - x ^{2}} + \frac{1}{- 5 - 6 x - x ^{2}}

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

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

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

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

Simplify

3 (- - x^{2} - 6 x - 5 - 3 arcsin (\frac{1}{2} (x + 3))) + arcsin (\frac{1}{2} (x + 3)) : - 3 - x^{2} - 6 x - 5 - 8 arcsin (\frac{1}{2} (x + 3))

= - 3 - x^{2} - 6 x - 5 - 8 arcsin (\frac{1}{2} (x + 3))

Add a constant to the solution

= - 3 - x^{2} - 6 x - 5 - 8 arcsin (\frac{1}{2} (x + 3)) + C

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What is the integral of (3x+1)/(sqrt(-5-6x-x^2)) ?
The integral of (3x+1)/(sqrt(-5-6x-x^2)) is -3sqrt(-x^2-6x-5)-8arcsin(1/2 (x+3))+C