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

The given integral,
$\begin{array}{l}\mathrm{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}+4{)}^{8/7}(\mathrm{x}-3{)}^{6/7}}\\ =\int \frac{\frac{\mathrm{dx}}{(\mathrm{x}+4{)}^2}}{\frac{(\mathrm{x}+4{)}^{8/7}(\mathrm{x}-3{)}^{6/7}}{(\mathrm{x}+4{)}^2}}=\int \frac{\frac{\mathrm{dx}}{(\mathrm{x}+4{)}^2}}{{\left(\frac{\mathrm{x}-3}{\mathrm{x}+4}\right)}^{6/7}}\end{array}$
Now, let $\frac{\mathrm{x}-3}{\mathrm{x}+4}={\mathrm{t}}^7$
$\begin{array}{l}\Rightarrow \frac{(\mathrm{x}+4)-(\mathrm{x}-3)}{(\mathrm{x}+4{)}^2}\mathrm{dx}=7{\mathrm{t}}^6\mathrm{dt}\\ \Rightarrow \frac{7}{(\mathrm{x}+4{)}^2}\mathrm{dx}=7{\mathrm{t}}^6\mathrm{dt}\end{array}$
So, $\mathrm{I}=\int \frac{{\mathrm{t}}^6\mathrm{dt}}{{\mathrm{t}}^6}=\int \mathrm{dt}=\mathrm{t}+\mathrm{C}$
$\therefore \mathrm{I}={\left(\frac{\mathrm{x}-3}{\mathrm{x}+4}\right)}^{1/7}+\mathrm{C}$