Question
$\int\limits^1_0\frac{\text{x}}{(1-\text{x})^{54}}\text{ dx}=$

  1. $\frac{15}{16}$

  2. $\frac{3}{16}$

  3. $-\frac{3}{16}$

  4. $-\frac{16}{3}$

Answer

  1. $-\frac{16}{3}$

solution:

$\text{I}=\int\limits^1_0\frac{\text{x}}{(1-\text{x})^{\frac{5}{4}}}\text{ dx}$

Put, 1 - x = t ⇒ x =1 -t

⇒ dx = -dt

x
0
1
t
1
0

$\text{I}=\int\limits^0_1\frac{(1-\text{t})(-\text{dt})}{\text{t}^{\frac{5}{4}}}$

$\text{I}=\int\limits^1_0\Big(\text{t}^\frac{5}{4}-\text{t}^\frac{-1}{4}\Big)\text{dt}$

$\text{I}=\Bigg[\frac{\text{t}^{-\frac{5}{4}}}{\frac{-1}{4}}-\frac{\text{t}^\frac{3}{4}}{\frac{3}{4}}\Bigg]^1_0$

$\text{I}=-4-\frac{4}{3}$

$\text{I}=\frac{-16}{3}$

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