What is the integral of ((x+2))/([4-(x-2)^2]^{3/2)} ?

The integral of ((x+2))/([4-(x-2)^2]^{3/2)} is 1/((-x^2+4x)^{1/2)}+(x-2)/((-x^2+4x)^{1/2)}+C

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integral of ((x+2))/([4-(x-2)^2]^{3/2)}

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

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

Solution steps

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

Expand

\frac{x + 2}{[ 4 - ( x - 2 ) ^{2} ] ^{\frac{3}{2}}} : \frac{x}{( - x ^{2} + 4 x ) ^{\frac{3}{2}}} + \frac{2}{( - x ^{2} + 4 x ) ^{\frac{3}{2}}}

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

\int \frac{x}{( - x ^{2} + 4 x ) ^{\frac{3}{2}}} d x = \frac{1}{( - x ^{2} + 4 x ) ^{\frac{1}{2}}} + \frac{x - 2}{2 ( - x ^{2} + 4 x ) ^{\frac{1}{2}}}

\int \frac{2}{( - x ^{2} + 4 x ) ^{\frac{3}{2}}} d x = \frac{x - 2}{2 ( - x ^{2} + 4 x ) ^{\frac{1}{2}}}

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of ((x+2))/([4-(x-2)^2]^{3/2)} ?
The integral of ((x+2))/([4-(x-2)^2]^{3/2)} is 1/((-x^2+4x)^{1/2)}+(x-2)/((-x^2+4x)^{1/2)}+C