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The value of the integral \int \frac{d x}{(x-2)^{7 / 8}(x+3)^{9 / 8}} is equal to 

Option: 1

\frac{8}{5}\left(\frac{x-2}{x+3}\right)^{\frac{1}{8}}+C


Option: 2

\frac{5}{8}\left(\frac{x-2}{x+3}\right)^{\frac{1}{8}}+C


Option: 3

\frac{5}{8}\left(\frac{x+3}{x-2}\right)^{\frac{1}{8}}+C


Option: 4

\frac{8}{5}\left(\frac{x+3}{x-2}\right)^{\frac{1}{8}}+C


Answers (1)

best_answer

(a) Given integral is 

               I=\int \frac{d x}{(x-2)^{7 / 8}(x+3)^{9 / 8}}

On substituting \frac{x-2}{x+3}=t, we get

\begin{aligned} & \frac{(x+3) \times 1-(x-2) \times 1}{(x+3)^2} d x=d t \Rightarrow \frac{d x}{(x+3)^2}=\frac{d t}{5} \\ \\& \therefore \quad I=\int \frac{(x+3)^{7 / 8} d x}{(x-2)^{7 / 8}(x+3)^{7 / 8}(x+3)^{9 / 8}} \\ \\& =\int\left(\frac{x+3}{x-2}\right)^{7 / 8} \frac{d x}{(x+3)^2} \\ \\& =\int\left(\frac{1}{t}\right)^{7 / 8} \frac{d t}{5}=\frac{1}{5} \int t^{-7 / 8} d t \\ \\& =\frac{1}{5}\left[\frac{t^{1 / 8}}{1 / 8}\right]+C=\frac{8}{5}\left(\frac{x-2}{x+3}\right)^{\frac{1}{8}}+C \\ & \end{aligned}

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