Official solution
\[\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}} \]
= \[\int {{{dx} \over {{{\left( {x + 4} \right)}^2}{{\left( {{{x - 3} \over {x + 4}}} \right)}^{{6 \over 7}}}}}} \]
Put \[{{{x - 3} \over {x + 4}}}\] = t
\[ \Rightarrow \] \[\left\{ {{{\left( {x + 4} \right) - \left( {x - 3} \right)} \over {{{\left( {x + 4} \right)}^2}}}} \right\}dx\] = dt
\[ \Rightarrow \] \[{{dx} \over {{{\left( {x + 4} \right)}^2}}} = {{dt} \over 7}\]
= \[{1 \over 7}\int {{{dt} \over {{{\left( t \right)}^{{6 \over 7}}}}}} \]
= \[{1 \over 7}\left( {{{{t^{{1 \over 7}}}} \over {{1 \over 7}}}} \right)\] + C
= \[{\left( {{{x - 3} \over {x + 4}}} \right)^{{1 \over 7}}}\] + C