Question

The integral $\int \frac{dx}{{(x+4)}^{{\displaystyle \frac{8}{7}}}{(x-3)}^{{\displaystyle \frac{6}{7}}}}$ is equal to

• A. $-{\left(\frac{x-3}{x+4}\right)}^{-\frac{1}{7}}+C$
• B. $\frac{1}{2}{\left(\frac{x-3}{x+4}\right)}^{\frac{3}{7}}+C$
• C. ${\left(\frac{x-3}{x+4}\right)}^{\frac{1}{7}}+C$
• D. $-\frac{1}{13}{\left(\frac{x-3}{x+4}\right)}^{-\frac{13}{7}}+C$

Answer 1

The correct option is C

${\left(\frac{x-3}{x+4}\right)}^{\frac{1}{7}}+C$

Explanation for the correct option:

Evaluate the integral

$$\begin{gathered} I=\int \frac{dx}{{(x+4)}^{{\displaystyle \frac{8}{7}}}{(x-3)}^{{\displaystyle \frac{6}{7}}}} \\ I=\int \frac{{(x+4)}^{{\displaystyle \frac{6}{7}}}dx}{{\left(x-3\right)}^{\frac{6}{7}}\times {(x+4)}^2} \\ =\int {\left(\frac{x-3}{x+4}\right)}^{-\frac{6}{7}}\times \frac{dx}{{(x+4)}^2} \end{gathered}$$

Put $\frac{x-3}{x+4}=t\Rightarrow dt=7\left(\frac{1}{{(x+4)}^2}\right)dx$

$$\begin{gathered} I=\frac{\int t^{-{\displaystyle \frac{6}{7}}}}{7}dt \\ =t^{\frac{1}{7}}+C \\ ={\left(\frac{x-3}{x+4}\right)}^{\frac{1}{7}}+C \end{gathered}$$

Hence, option (C) is correct.