Find: (3x + 5)5+4x-2x^2dx - Vidyadip

Question

Find: $\int(\text{3x + 5)}\sqrt{5+4x-2x^2}\text{dx}$

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Answer

$\text{(3x + 5)}\sqrt{5 + \text{4x} - \text{2x}^{2}} \text{ dx}$
$\text{let 3x + 5 = A (4 - 4x) + B}$
$\Rightarrow\text{A} = -\frac{3}{4},\text{B} = 8$
$\text{I} = -\frac{3}{4}(4 - \text{4x)}\sqrt{5 + \text{4x - 2x}^{2}} \text{ dx} + 8 \sqrt{5 + 4\text{x - 2x}^{2}}\text{ dx}$
$= -\frac{3}{4}\text{I}_{1} + 8\text{I}_{2}(\text{let)}$
$\text{For I}_{1}, \text{put} 5 + \text{4x - 2x}^{2} = \text{t}$
$\Rightarrow(4 - \text{4x) dx = dt}$
$-\frac{3}{4}\text{I}_{1} = -\frac{3}{4}\sqrt{\text{t}}\text{ dt} = -\frac{3}{4}\times\frac{2}{4}\text{t}^{3/2}$
$= -\frac{1}{2}(5 + \text{4x}{ - }{{2x}^{2})^{3/2}}$
$\text{8I}_{2} = 8\sqrt{2}\sqrt{\frac{7}{2} - \text({x - 1)}^{2}}\text{ dx}$
$ = 8\sqrt{2}\bigg[\frac{\text{x - 1}}{2}\sqrt{\frac{5}{2} + \text{2x - x}^{2}} + \frac{7}{4}\sin^{-1} \frac{\sqrt{2}(\text{x - 1)}}{\sqrt{7}}\bigg]$
$\text{I} =- \frac{1}{2}(5 + \text{4x - 2x}^{2})^{3/2} + 4\sqrt{2}(\text{x - 1)}{\sqrt{\frac{5}{2} + \text{2x - x}^{2}}} + 14\sqrt{2}\sin^{-1}\frac{\sqrt{2}\text{(x - 1)}}{\sqrt{7}} + \text{C}$

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