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 $$k$$.


ANSWER

7


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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