Question 971771
Find an antiderivative for {{{-5(-4x^2-1)^2}}} 
<pre>
{{{-5(-4x^2-1)^2}}}{{{""=""}}}{{{-5(16x^4+8x^2+1)}}}{{{""=""}}}{{{-80x^4-40x^2-5}}}

Since that has 3 terms, 

Let an antiderivative be = {{{a*x^n+bx^m+cx^p}}}

the derivative of {{{a*x^n+bx^m+cx^p}}} is {{{a*n*x^(n-1)+b*m*x^(m-1)+c*p*x^(p-1)}}}

Therefore {{{a*n*x^(n-1)+b*m*x^(m-1)+c*p*x^(p-1)}}}{{{""=""}}}{{{-80x^4-40x^2-5}}}

Write the 5 as {{{5*x^0}}}

{{{a*n*x^(n-1)+b*m*x^(m-1)+c*p*x^(p-1)}}}{{{""=""}}}{{{-80x^4-40x^2-5x^0}}}

Set the corresponding exponents equal to each other

{{{n-1=4}}}, {{{m-1=2}}}, {{{p-1=0}}}

Solve for the exponents:

{{{n=5}}}, {{{m=3}}}, {{{p=1}}}

Substituting,

{{{a*5*x^4+b*3*x^2+c*1*x^0}}}{{{""=""}}}{{{-80x^4-40x^2-5x^0}}}

{{{5ax^4+3bx^2+cx^0}}}{{{""=""}}}{{{-80x^4-40x^2-5x^0}}}

Set the coefficients of like powers of x equal to each other

{{{5a=-80}}}, {{{3b=-40}}}, {{{c=-5}}}

{{{a=-16}}}, {{{b=-40/3}}}, {{{c=-5}}}

Substituting those and {{{n=5}}}, {{{m=3}}}, {{{p=1}}}

So an antiderivative = {{{a*x^n+bx^m+cx^p}}}{{{""=""}}}{{{-16x^5-expr(40/3)x^3-5x^1}}}

Answer:  {{{-16x^5-expr(40/3)x^3-5x}}}

Edwin</pre>