Evaluate `int (x+3)sqrt(3-4x-x^2)dx`

Explanation

`L e t I=int(x+3) sqrt{3-4 x-x^{2}} d x`

We express `x+3=A(frac{d}{d x}(3-4 x-x^{2}))+B` `x+3=A(-4-2 x)+B`

Equating coefficients

`1=-2 A and 3=-4 A+B`
or `A=-frac{1}{2}` and `B=1`

`therefore I=int(-frac{1}{2}(-4-2 x)+1) sqrt{3-4 x-} `

`=-frac{1}{2} int(-4-2 x) sqrt{3-4 x-x^{2}} {dx}+int `

`=-frac{1}{2} I_{1}+I_{2 ldots}(1) `

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