What is the integral of ((3+4x))/(sqrt(3x-x^2-2)) ?

The integral of ((3+4x))/(sqrt(3x-x^2-2)) is 9arcsin(2x-3)-4sqrt(-x^2+3x-2)+C

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integral of ((3+4x))/(sqrt(3x-x^2-2))

\int \frac{( 3 + 4 x )}{3 x - x ^{2} - 2} d x

9 arcsin (2 x - 3) - 4 - x^{2} + 3 x - 2 + C

Solution steps

\int \frac{3 + 4 x}{3 x - x ^{2} - 2} d x

Expand

\frac{3 + 4 x}{3 x - x ^{2} - 2} : \frac{3}{3 x - x ^{2} - 2} + \frac{4 x}{3 x - x ^{2} - 2}

= \int \frac{3}{3 x - x ^{2} - 2} + \frac{4 x}{3 x - x ^{2} - 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}{3 x - x ^{2} - 2} d x + \int \frac{4 x}{3 x - x ^{2} - 2} d x

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

\int \frac{4 x}{3 x - x ^{2} - 2} d x = 4 (- - x^{2} + 3 x - 2 + \frac{3}{2} arcsin (2 x - 3))

= 3 arcsin (2 x - 3) + 4 (- - x^{2} + 3 x - 2 + \frac{3}{2} arcsin (2 x - 3))

Simplify

3 arcsin (2 x - 3) + 4 (- - x^{2} + 3 x - 2 + \frac{3}{2} arcsin (2 x - 3)) : 9 arcsin (2 x - 3) - 4 - x^{2} + 3 x - 2

= 9 arcsin (2 x - 3) - 4 - x^{2} + 3 x - 2

Add a constant to the solution

= 9 arcsin (2 x - 3) - 4 - x^{2} + 3 x - 2 + C

x \to 3 lim (\frac{x + 3}{x + 6})

\int \frac{1}{t ^{2} - 10 t + 41} d t

f (x) = x^{2} arctan (5 x)

\int \frac{x + 5}{( x + 2 ) ( x - 1 )} d x

y = 5 + 5 x^{2} - 2 x^{3}, (1, 8)

What is the integral of ((3+4x))/(sqrt(3x-x^2-2)) ?
The integral of ((3+4x))/(sqrt(3x-x^2-2)) is 9arcsin(2x-3)-4sqrt(-x^2+3x-2)+C