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

The integral of (x+9)/(sqrt(9-3x-2x^2)) is 1/(4sqrt(2))(-sqrt(-16x^2-24x+72)+33arcsin(4/9 x+1/3))+C

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integral of (x+9)/(sqrt(9-3x-2x^2))

\int \frac{x + 9}{9 - 3 x - 2 x ^{2}} d x

\frac{1}{4 2} (- - 16 x^{2} - 24 x + 72 + 33 arcsin (\frac{4}{9} x + \frac{1}{3})) + C

Solution steps

\int \frac{x + 9}{9 - 3 x - 2 x ^{2}} d x

Complete the square

9 - 3 x - 2 x^{2} : - 2 (x + \frac{3}{4})^{2} + \frac{81}{8}

= \int \frac{x + 9}{- 2 ( x + \frac{3}{4} ) ^{2} + \frac{81}{8}} d x

Apply u-substitution

: \int \frac{4 u + 33}{2 - 16 u ^{2} + 81} d u

= \int \frac{4 u + 33}{2 - 16 u ^{2} + 81} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{4 u + 33}{- 16 u ^{2} + 81} d u

Expand

\frac{4 u + 33}{- 16 u ^{2} + 81} : \frac{4 u}{- 16 u ^{2} + 81} + \frac{33}{- 16 u ^{2} + 81}

= \frac{1}{2} \cdot \int \frac{4 u}{- 16 u ^{2} + 81} + \frac{33}{- 16 u ^{2} + 81} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{4 u}{- 16 u ^{2} + 81} d u + \int \frac{33}{- 16 u ^{2} + 81} d u)

\int \frac{4 u}{- 16 u ^{2} + 81} d u = - \frac{1}{4} - 16 u^{2} + 81

\int \frac{33}{- 16 u ^{2} + 81} d u = \frac{33}{4} arcsin (\frac{4}{9} u)

= \frac{1}{2} (- \frac{1}{4} - 16 u^{2} + 81 + \frac{33}{4} arcsin (\frac{4}{9} u))

Substitute back

u = x + \frac{3}{4}

= \frac{1}{2} - \frac{1}{4} - 16 (x + \frac{3}{4})^{2} + 81 + \frac{33}{4} arcsin (\frac{4}{9} (x + \frac{3}{4}))

Simplify

\frac{1}{2} - \frac{1}{4} - 16 (x + \frac{3}{4})^{2} + 81 + \frac{33}{4} arcsin (\frac{4}{9} (x + \frac{3}{4})) : \frac{1}{4 2} (- - 16 x^{2} - 24 x + 72 + 33 arcsin (\frac{4}{9} x + \frac{1}{3}))

= \frac{1}{4 2} (- - 16 x^{2} - 24 x + 72 + 33 arcsin (\frac{4}{9} x + \frac{1}{3}))

Add a constant to the solution

= \frac{1}{4 2} (- - 16 x^{2} - 24 x + 72 + 33 arcsin (\frac{4}{9} x + \frac{1}{3})) + C

x \to 0 lim ((1 + \frac{3}{x ^{2}})^{x})

y^{^{''}} + b^{2} y = 0

y^{^{'}} = \frac{( e ^{x - y} )}{1 + e ^{x}}

\int \frac{24}{x ^{2}} d x

\frac{d y}{d x} + 2 x y^{5} = 0

What is the integral of (x+9)/(sqrt(9-3x-2x^2)) ?
The integral of (x+9)/(sqrt(9-3x-2x^2)) is 1/(4sqrt(2))(-sqrt(-16x^2-24x+72)+33arcsin(4/9 x+1/3))+C