Question

The integral $\int \frac{dx}{(x+4{)}^{\frac{8}{7}}(x-3{)}^{\frac{6}{7}}}$ is equal to : (where $C$ is a constant of intergration)

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

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

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

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