Question 964019
{{{(2ax+x^2)expr(dy/dx)}}}{{{""=""}}}{{{a^2+2ax}}}

{{{dy}}}{{{""=""}}}{{{expr((a^2+2ax)/(2ax+x^2))dx}}}

{{{int(dy)}}}{{{""=""}}}{{{int(expr((a^2+2ax)/(2ax+x^2))dx)}}}

{{{y}}}{{{""=""}}}{{{int(expr((a(a+2x))/(2ax+x^2))dx)}}}

{{{y}}}{{{""=""}}}{{{a*int(expr((a+2x)/(2ax+x^2))dx)}}}

If the numerator were the derivative of the denominator, we
would be able to integrate using {{{int(du/u)=ln(abs(u))+C}}}

The derivative of the denominator is 2a+2x.
The numerator a+2x only needs to have "a" added to it to become that
So in the numerator we will add "a" and subtract "a"

{{{y}}}{{{""=""}}}{{{a*int(expr((red(a)+a+2x-red(a))/(2ax+x^2))dx)}}} 


{{{y}}}{{{""=""}}}{{{a*int(expr((2a+2x-red(a))/(2ax+x^2))dx)}}} 

Write as the difference of two fractions and integrate each:

{{{y}}}{{{""=""}}}{{{a*int((

expr((2a+2x)/(2ax+x^2)) - expr(a/(2ax+x^2)))dx )}}}

{{{y}}}{{{""=""}}}{{{a*int(((

expr((2a+2x)/(2ax+x^2)))^"" - (expr(a/(2ax+x^2))))dx )}}}

{{{y}}}{{{""=""}}}{{{a*int(expr((2a+2x)/(2ax+x^2))dx)}}}{{{""-""}}}{{{a*int(expr(a/(2ax+x^2))dx)}}}

Use the natural log formula on the first integral, and take the 
constant "a" outside the second integral:

{{{y}}}{{{""=""}}}{{{a*ln(abs(2ax+x^2))}}}{{{""-""}}}{{{a^2*int(expr(1/(2ax+x^2))dx)}}}

To the side, let's complete the square in the denominator by
adding and subtrracting a<sup>2</sup>:

      {{{2ax+x^2}}}{{{""=""}}}{{{x^2+2ax+a^2-a^2}}}{{{""=""}}}{{{(x+a)^2-a^2}}}

{{{y}}}{{{""=""}}}{{{a*ln(abs(2ax+x^2))}}}{{{""-""}}}{{{a^2*int(expr(1/((x+a)^2-a^2))dx)}}}

To do the remaining integral we use the formula {{{int( du/(u^2-a^2) )}}}{{{""=""}}}{{{expr( 1/(2a) )ln(abs( (u-a)/(u+a)) )+C}}}

{{{y}}}{{{""=""}}}{{{a*ln(abs(2ax+x^2))}}}{{{""-""}}}{{{a^2*expr( 1/(2a) )ln(abs( ((x+a)-a)/((x+a)+a)) )+C}}}

{{{y}}}{{{""=""}}}{{{a*ln(abs(2ax+x^2))}}}{{{""-""}}}{{{expr( a/2 )ln(abs( x/(x+2a)) )+C}}}

Edwin</pre>