Let I= (1-x)^ 23 dx Substituting 1 - x = t and dx = -dt, we get I= (1-t)^ 23 dt =- (t^ 23 -t^ 24 )dt =- ( t^ 24 24 - t^ 25 25 )+C = t^ 25 24 - t^ 24 25 +C = (1-x)^ 25 25 - (1-x)^ 24 24 +C I= (1-x)^ 25 25 - (1-x)^ 24 24 +C =1 600 (1-x)^ 24 [24(1-x)-25 ]+C =1 600 (1-x)^ 24 [24-24x-25 ]+C =1 600 (1-x)^ 24 [-1-24x ]+C =1 600 (1-x)^ 24 - [1+24x ]+C =1 600 (1-x)^ 24 (1+24x)+C

(1-x)^ 23 dx

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Question

$\int\text{x}(1-\text{x})^{23}\text{dx}$

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Answer

Let $\text{I}=\int\text{x}(1-\text{x})^{23}\text{dx}$
Substituting 1 - x = t and dx = -dt, we get
$\text{I}=\int(1-\text{t})^{23}\text{dt}$
$=-\int(\text{t}^{23}-\text{t}^{24})\text{dt}$
$=-\int\Big(\frac{\text{t}^{24}}{24}-\frac{\text{t}^{25}}{25}\Big)+\text{C}$
$=\frac{\text{t}^{25}}{24}-\frac{\text{t}^{24}}{25}+\text{C}$
$=\frac{(1-\text{x})^{25}}{25}-\frac{(1-\text{x})^{24}}{24}+\text{C}$
$\therefore\ \text{I}=\frac{(1-\text{x})^{25}}{25}-\frac{(1-\text{x})^{24}}{24}+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[24(1-\text{x})-25\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[24-24\text{x}-25\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[-1-24\text{x}\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\times-\big[1+24\text{x}\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}(1+24\text{x})+\text{C}$

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