Question 474425
{{{(x^2 - 8x + 15)expr(dy/dx) = 6y + 5}}}
<pre>
Multiply through by dx

{{{(x^2 - 8x + 15)dy = (6y + 5)dx}}}

Separate the variables:

{{{dy/(6y+5) = dx/(x^2-8x+15)}}}

Integrating the left side using formula {{{int(du/u)}}} = {{{ln(abs(u))+C}}}

{{{int(dy/(6y+5))}}} = {{{expr(1/red(6))int(red(6)dy/(6y+5))}}} = {{{expr(1/6)ln(6y+5)+C[1]}}}

To integerate the right side, we must first complete the
square on the denominator:

x² - 8x + 15

multiply -8 by 1/2, get -4, then square, getting +16
Add 16 then subtract 16

x² - 8x + 16 - 16 + 15

Factor first three terms as square of a binomian, combine last two terms:

(x - 4)² - 1

{{{int(dx/(x^2-8x+15))}}} = {{{int(dx/((x-4)^2-1))}}}

We use formula {{{int(du/(u^2-a^2))}}} = {{{expr(1/(2a))ln(abs((u-a)/(u+a)))+C}}}

{{{int(dx/((x-4)^2-1))}}} = {{{expr(1/(2*1))ln(abs(((x-4)-1)/((x-4)+1)))+C[2]}}} = {{{expr(1/2)ln(abs((x-5)/(x-3)))+C[2]}}}

So we have:

{{{expr(1/6)ln(6y+5)+C[1]}}} = {{{expr(1/2)ln(abs((x-5)/(x-3)))+C[2]}}}

Let arbitrary constant {{{C[3] = C[2]-C[1]}}}

{{{expr(1/6)ln(6y+5)}}} = {{{expr(1/2)ln(abs((x-5)/(x-3)))+C[3]}}}

Multiply through by 6

{{{ln(6y+5)}}} = {{{3ln(abs((x-5)/(x-3)))+3C[3]}}}

Use the coefficient/exponent rule of logarithms and 
write arbitrary constant {{{3C[3]}}} as {{{ln(C)}}},
choosing C > 0.

{{{ln(6y+5)}}} = {{{ln(abs((x-5)/(x-3))^3)+ln(C)}}}

Use the sum/product rule of logarithms on the right:

{{{ln(6y+5)}}} = {{{ln(C*abs((x-5)/(x-3))^3)}}}


{{{6y+5}}} = {{{C*abs((x-5)/(x-3))^3}}}

If the arbitrary constant is chosen properly we may 
erase the absolute values:

{{{6y+5}}} = {{{C*expr((x-5)/(x-3)^3)}}}

or if we like:

{{{(6y+5)(x-3)^3)}}} = {{{C(x-5)}}}

Edwin</pre>