Mathematics

# $\displaystyle \int \frac{dx}{\left ( x+\alpha \right )^{8/7}\left ( x-\beta \right )^{6/7}}=\frac{k}{\alpha +\beta }\left ( \frac{x-\beta }{x+\alpha } \right )^{1/k}.$ Find the value of $k$.

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##### SOLUTION
Let $\displaystyle I=\int \dfrac { dx }{ \left( x+\alpha \right) ^{ 8/7 }\left( x-\beta \right) ^{ 6/7 } }$

$\displaystyle I=\dfrac{dx}{\left ( x+\alpha \right )^{8/7+6/7}\cdot \left ( \dfrac{x-\beta }{x+\alpha } \right )^{8/7}}$

$\displaystyle =\dfrac{dx}{\left ( x+\alpha \right )^{2}\left [ 1-\dfrac{\alpha +\beta }{x+\alpha } \right ]^{6/7}}$

Substitute $\displaystyle 1-\dfrac{\alpha +\beta }{x+\alpha }=t\therefore \dfrac{\alpha +\beta}{\left ( x+\alpha \right )^{2}}dx=dt$

$\displaystyle \therefore I=\dfrac{1}{\left ( \alpha +\beta \right )}\int \dfrac{dt}{t^{6/7}}=\dfrac{7}{\alpha +\beta }t^{1/7}$

$\displaystyle =\dfrac{7}{\alpha +\beta }\left [ 1-\dfrac{\alpha +\beta }{x+\alpha } \right ]^{1/7}=\dfrac{7}{\alpha +\beta }\left ( \dfrac{x-\beta }{x+\alpha } \right )^{1/7}$

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One Word Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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