Mathematics

# $\int_{}^{} {\dfrac{{{x^2}}}{{\sqrt {1 - x} }}dx = \int_{}^{} {\dfrac{{{{\left( {1 - u} \right)}^2}}}{{\sqrt u }}\left( { - du} \right)} }$

$\frac{2}{3}{\left( {1 - x} \right)^{\frac{3}{2}}}\left( {3{x^2} + 4x + 5} \right)$

##### SOLUTION
$\int_{}^{} {\cfrac{{{x^2}}}{{\sqrt {1 - x} }}dx = \int_{}^{} {\cfrac{{{{\left( {1 - u} \right)}^2}}}{{\sqrt u }}\left( { - du} \right)} }$

Let $u=1-x$ $\Rightarrow$ du=-dx
$= - \int_{}^{} {\cfrac{{1 - 2u + {u^2}}}{{\sqrt u }}du}$
=$- \int_{}^{} {\cfrac{{du}}{{\sqrt u }} + 2\int_{}^{} {\sqrt u } du - \int_{}^{} {{u^{\cfrac{3}{2}}}du} }$
=$- 2{u^{\cfrac{1}{2}}} + 2.\cfrac{2}{3}{u^{\cfrac{3}{2}}} - \cfrac{2}{5}{u^{\cfrac{5}{2}}} + C$
=$- 2{u^{\cfrac{1}{2}}}\left[ {1 - \cfrac{2}{3}u + \cfrac{1}{5}{u^2}} \right] + C$
=$\cfrac{{ - 2}}{{15}}{u^{\cfrac{1}{2}}}\left[ {15 - 10\left( {1 - x} \right) + 3{{\left( {1 - x} \right)}^2}} \right] + C$
=$\cfrac{{ - 2}}{{15}}\left( {\sqrt {1 - x} } \right)\left[ {3{x^2} + 4x + 8} \right] + C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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