What is the integral of x/((sqrt(5-4x-x^2))^3) ?

The integral of x/((sqrt(5-4x-x^2))^3) is 1/(sqrt(-x^2-4x+5))-(2(x+2))/(9sqrt(-x^2-4x+5))+C

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integral of x/((sqrt(5-4x-x^2))^3)

\int \frac{x}{( 5 - 4 x - x ^{2} ) ^{3}} d x

\frac{1}{- x ^{2} - 4 x + 5} - \frac{2 ( x + 2 )}{9 - x ^{2} - 4 x + 5} + C

Solution steps

\int \frac{x}{( 5 - 4 x - x ^{2} ) ^{3}} d x

Complete the square

5 - 4 x - x^{2} : - (x + 2)^{2} + 9

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

Apply u-substitution

: \int \frac{u - 2}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u

= \int \frac{u - 2}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u

Expand

\frac{u - 2}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} : \frac{u}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} - \frac{2}{( - u ^{2} + 9 ) ^{\frac{3}{2}}}

= \int \frac{u}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} - \frac{2}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} 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

= \int \frac{u}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u - \int \frac{2}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u

\int \frac{u}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u = \frac{1}{- u ^{2} + 9}

\int \frac{2}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u = \frac{2 u}{9 9 - u ^{2}}

= \frac{1}{- u ^{2} + 9} - \frac{2 u}{9 9 - u ^{2}}

Substitute back

u = x + 2

= \frac{1}{- ( x + 2 ) ^{2} + 9} - \frac{2 ( x + 2 )}{9 9 - ( x + 2 ) ^{2}}

Simplify

\frac{1}{- ( x + 2 ) ^{2} + 9} - \frac{2 ( x + 2 )}{9 9 - ( x + 2 ) ^{2}} : \frac{1}{- x ^{2} - 4 x + 5} - \frac{2 ( x + 2 )}{9 - x ^{2} - 4 x + 5}

= \frac{1}{- x ^{2} - 4 x + 5} - \frac{2 ( x + 2 )}{9 - x ^{2} - 4 x + 5}

Add a constant to the solution

= \frac{1}{- x ^{2} - 4 x + 5} - \frac{2 ( x + 2 )}{9 - x ^{2} - 4 x + 5} + C

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What is the integral of x/((sqrt(5-4x-x^2))^3) ?
The integral of x/((sqrt(5-4x-x^2))^3) is 1/(sqrt(-x^2-4x+5))-(2(x+2))/(9sqrt(-x^2-4x+5))+C