What is the integral of r/((5-4r-r^2)^{3/2)} ?

The integral of r/((5-4r-r^2)^{3/2)} is 1/((-r^2-4r+5)^{1/2)}-(2(r+2))/(9(-r^2-4r+5)^{1/2)}+C

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integral of r/((5-4r-r^2)^{3/2)}

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

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

Solution steps

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

Complete the square

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

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

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 ) ^{\frac{1}{2}}}

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

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

Substitute back

u = r + 2

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

Simplify

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

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

Add a constant to the solution

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

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