(1-x)^ 23 dx - Vidyadip

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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