Question 985388
<pre>
Rather than do your problem for you, I will instead do one 
EXACTLY like yours, step by step.  Instead of yours I will 
do this one:

{{{int((5x+7)/(3x^2+11x-4),dx)}}}

We factor the denominator and set up for partial fractions:

{{{(5x+7)/((3x-1)(x+4))}}}{{{""=""}}}}{{{A/(3x-1)+B/(x+4)}}}

Clear fractions:

{{{5x+7}}}{{{""=""}}}{{{A(x+4)+B(3x-1)}}}

{{{5x+7}}}{{{""=""}}}{{{Ax+4A+3Bx-B}}}

Equate the coefficients of x:

{{{5}}}{{{""=""}}}{{{A+3B}}}

Equate the constant terms

{{{7}}}{{{""=""}}}{{{4A-B}}}

Solve the system of equations by substitution:

{{{system(A+3B=5,4A-B=7)}}}

A=5-3B,       4A-B=7
         4(5-3B)-B=7 
          20-12B-B=7
            20-13B=7
              -13B=-13
                 B=1

A=5-3(1)
A=5-3
A=2

{{{(5x+7)/((3x-1)(x+4))}}}{{{""=""}}}}{{{2/(3x-1)+1/(x+4)}}}

{{{int((5x+7)/(3x^2+11x-3),dx)}}}{{{""=""}}}

{{{int((2/(3x-1)+1/(x+4)),dx)}}}{{{""=""}}}{{{int((2/(3x-1)),dx)}}}{{{""+""}}}{{{int((1/(x+4)),dx)}}}{{{""=""}}}

{{{2int((1/(3x-1)),dx)}}}{{{""+""}}}{{{ln(x+4)}}}{{{""+""}}}{{{C}}}{{{""=""}}}

{{{expr(2/3)int((3/(3x-1)),dx)}}}{{{""+""}}}{{{ln(x+4)}}}{{{""+""}}}{{{C}}}{{{""=""}}} {{{expr(2/3)ln(3x-1)}}}{{{""+""}}}{{{ln(x+4)}}}{{{""+""}}}{{{C}}}

Edwin</pre>