2-3x 1+3x dx - Vidyadip

Question

$\int\frac{2-3\text{x}}{\sqrt{1+3\text{x}}}\text{dx}$

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Answer

Let I $=\int\frac{2-3\text{x}}{\sqrt{1+3\text{x}}}\times\text{dx}.$ Then,
$\text{I}=\int\frac{2-3\text{x}-1+1}{\sqrt{1+3\text{x}}}\text{dx}$
$=\int\frac{-3\text{x}-1+3}{\sqrt{1+3\text{x}}}\text{dx}$
$=\int-\frac{(3\text{x}+1)}{\sqrt{1+3\text{x}}}\text{dx}+3\int\frac{1}{\sqrt{1+3\text{x}}}\text{dx}$
$=-1\int\frac{1+3\text{x}}{\sqrt{1+3\text{x}}}\text{dx}+3\int\frac{1}{\sqrt{1+3\text{x}}}\text{dx}$
$=-1\int(1+3\text{x})^\frac{1}{2}\text{dx}+3\int(1+3\text{x})^\frac{-1}{2}\text{dx}$
$=-1\times\frac{(1+3\text{x})^\frac{3}{2}}{\frac{3}{2}\times3}+3\times\frac{(1+3\text{x})^\frac{1}{2}}{\frac{1}{2}\times3}+\text{C}$
$=-\frac{2}{9}\times(1+3\text{x})^\frac{3}{2}+2(1+3\text{x})^\frac{1}{2}+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[-\frac{1}{9}(1+3\text{x})^1+1\Big]+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[\frac{-1-3\text{x}+9}{9}\Big]+\text{C}$
$=2(1+3\text{x})^\frac{1}{2}\Big[\frac{8-3\text{x}}{9}\Big]+\text{C}$
$=\frac{2}{9}\sqrt{1+3\text{x}}(8-3\text{x})+\text{C}$
$\therefore\text{I}=\frac{2}{9}(8-9\text{x})\sqrt{1+3\text{x}}+\text{C}$

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