x^2 x-1 dx - Vidyadip

Question

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

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Answer

Let $\text{I}=\int\frac{\text{x}^2}{\sqrt{\text{x}-1}}\text{dx}$
Substituting x - 1 = t and dx = dt we get
$\text{I}=\int\frac{(\text{t}+1)^2}{\sqrt{\text{t}}}\text{dx}$
$=\int\frac{\text{t}^2+1+2\text{t}}{\sqrt{\text{t}}}\text{dt}$
$=\int\Big(\text{t}^\frac{3}{2}+\text{t}^\frac{-1}{2}+2\text{t}^\frac{-1}{2}\Big)\text{dt}$
$=\frac{2}{5}\text{t}^\frac{5}{2}+2\text{t}^\frac{1}{2}+\frac{4}{3}\text{t}^\frac{3}{2}+\text{C}$
$=\frac{6\text{t}^\frac{5}{2}+30\text{t}^\frac{1}{2}+20\text{t}^\frac{3}{2}}{15}+\text{C}$
$=\frac{2}{5}\text{t}^\frac{1}{2}\big(3\text{t}^2+15+10\text{t}\big)+\text{C}$
$=\frac{2}{5}\sqrt{\text{x}-1}\Big(3\big(\text{x}-1\big)^2+15+10(\text{x}-1)\Big)+\text{C}$
$=\frac{2}{5}\sqrt{\text{x}-1}\Big(3\big(\text{x}^2+1-2\text{x}\big)+15+10\text{x}-10\Big)+\text{C}$
$=\frac{2}{5}\sqrt{\text{x}-1}\big(3\text{x}^2+4\text{x}+8\big)+\text{C}$
$\therefore\ \text{I}=\frac{2}{5}\big(3\text{x}^2+4\text{x}+8\big)+\sqrt{\text{x}-1}+\text{C}$

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