Question
$\int(\text{x}+2)\sqrt{3\text{x}+5}\text{ dx}$

Answer

$\text{Let I}=\int(\text{x}+2)\sqrt{3\text{x}+5}\text{ dx}$
$\text{Putting}\ 3\text{x}+5=\text{t}$
$\Rightarrow\text{x}=\frac{\text{t}-5}{3}$
$\Rightarrow3\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{3}$
$\therefore\text{I}=\int\Big(\frac{\text{t}-5}{3}+2\Big)\sqrt{\text{t}}\frac{\text{dt}}{3}$
$=\frac{1}{3}\int\Big(\frac{\text{t}-5+6}{3}\Big)\sqrt{t}\text{ dt}$
$=\frac{1}{9}\int(\text{t}+1)\sqrt{\text{t}}\text{ dt}$
$=\frac{1}{9}\int\Big(\text{t}^\frac{3}{2}+\text{t}^\frac{1}{2}\Big)\text{dt}$
$=\frac{1}{9}\bigg[\frac{2}{5}(3\text{x}+5)^\frac{5}{2}+\frac{2}{3}(3\text{x}+5)^\frac{3}{2}\bigg]+\text{C}$ $[\because\text{t}=3\text{x}+5]$
$=\frac{2}{9}\bigg[(3\text{x}+5)^\frac{3}{2}\Big\{\frac{3\text{x}+5}{5}+\frac{1}{3}\Big\}\bigg]+\text{C}$
$=\frac{2}{9}\bigg[(3\text{x}+5)^\frac{3}{2}\Big\{\frac{9\text{x}+15+5}{15}\Big\}\bigg]+\text{C}$
$=\frac{2}{9}\bigg[(3\text{x}+5)^\frac{3}{2}\Big\{\frac{9\text{x}+20}{15}\Big\}\bigg]+\text{C}$
$=\frac{2}{135}(3\text{x}+5)^\frac{3}{2}(9\text{x}+20)+\text{C}$

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