x^2 1-x dx Let 1-x=t =1-t 1=- dt dx =-dt Now, x^2 1-x dx = (1-t)^2 t dt = ( 1-t^2-2t t )dt = (1 t + t^2 t - 2t t )dt = (t^ -1 2 +t^ 3 2 -2t^ 1 2 )dt = [ t^ -1 2 +1 -1 2 +1 + t^ 3 2 +1 3 2 +1 - 2t^ 1 2 +1 1 2 +1 ]+C =2t^ 1 2 +2 5 t^ 5 2 -4 3 t^ 3 2 +C =2t^ 1 2 [1+ t^2 5 -2 3 t ]+C =2t^ 1 2 [ 15+3t^2-10t 15 ]+C =2 1-x [ 15+3(1-x)^2-10(1-x) 15 ]+C =2 15 1-x [15+3(1^2+x^2-2x)-10+10x ]+C =2 15 1-x [15+3+3x^2-6x+10+10x ]+C =2 15 1-x [3x^2+4x+8 ]+C

x^2 1-x dx

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Question

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

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Answer

$\int\frac{\text{x}^2}{\sqrt{1-\text{x}}}\text{dx}$
Let $1-\text{x}=\text{t}$
$\Rightarrow\text{x}=1-\text{t}$
$\Rightarrow1=-\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=-\text{dt}$
Now, $\int\frac{\text{x}^2}{\sqrt{1-\text{x}}}\text{dx}$
$=\int\frac{(1-\text{t})^2}{\sqrt{\text{t}}}\text{dt}$
$=\int\Big(\frac{1-\text{t}^2-2\text{t}}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\frac{1}{\sqrt{\text{t}}}+\frac{\text{t}^2}{\sqrt{\text{t}}}-\frac{2\text{t}}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\text{t}^{-\frac{1}{2}}+\text{t}^{\frac{3}{2}}-2\text{t}^{\frac{1}{2}}\Big)\text{dt}$
$=\Bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}-\frac{2\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\text{t}^{\frac{1}{2}}+\frac{2}{5}\text{t}^{\frac{5}{2}}-\frac{4}{3}\text{t}^{\frac{3}{2}}+\text{C}$
$=2\text{t}^{\frac{1}{2}}\Big[1+\frac{\text{t}^2}{5}-\frac{2}{3}\text{t}\Big]+\text{C}$
$=2\text{t}^{\frac{1}{2}}\Big[\frac{15+3\text{t}^2-10\text{t}}{15}\Big]+\text{C}$
$=2\sqrt{1-\text{x}}\Big[\frac{15+3(1-\text{x})^2-10(1-\text{x})}{15}\Big]+\text{C}$
$=\frac{2}{15}\sqrt{1-\text{x}}\big[15+3(1^2+\text{x}^2-2\text{x})-10+10\text{x}\big]+\text{C}$
$=\frac{2}{15}\sqrt{1-\text{x}}\big[15+3+3\text{x}^2-6\text{x}+10+10\text{x}\big]+\text{C}$
$=\frac{2}{15}\sqrt{1-\text{x}}\big[3\text{x}^2+4\text{x}+8\big]+\text{C}$