Question 971772
Find an antiderivative for {{{-7(-6x-5^"")^(1/8)}}} 
<pre>
Let an antiderivative be = {{{k*u^n}}}

the derivative of {{{k*u^n}}} is {{{k*n*u^(n-1)expr((du)/(dx))}}}

Therefore {{{k*n*u^(n-1)expr((du)/(dx))}}}{{{""=""}}}{{{-7(-6x-5^"")^(1/8)}}}

Set the bases equal to each other and the exponents equal to each other

{{{u=(-6x-5^"")}}} and {{{n-1=1/8}}}

Take the derivative; solve for n

{{{(du)/(dx)=-6}}} and {{{n=1+1/8=9/8}}}

Substituting,

{{{k*expr(9/8)*(-6x-5^"")^(1/8)*(-6)}}}{{{""=""}}}{{{-7(-6x-5^"")^(1/8)}}}

Multiply both sides by 8 to clear the fraction:

 {{{k*9*(-6x-5^"")^(1/8)*(-6)}}}{{{""=""}}}{{{-56(-6x-5^"")^(1/8)}}}

{{{-54*k*(-6x-5^"")^(1/8)}}}{{{""=""}}}{{{-56(-6x-5^"")^(1/8)}}}

Divide both sides by {{{(-6x-5^"")^(1/8)}}}

{{{-54*k}}}{{{""=""}}}{{{-56}}}

{{{k}}}{{{""=""}}}{{{(-56)/(-54)}}}

{{{k}}}{{{""=""}}}{{{28/27}}}

So an antiderivative = {{{k*u^n}}}{{{""=""}}}{{{expr(28/27)(-6x-5^"")^(9/8)}}}

Edwin</pre>