Question 767757
solve the initial value problem:

please help with this question 
<pre>
{{{(dy)/(dx)}}}{{{""=""}}}{{{(e^(4x)+1)/(e^(4x)+4x+4)}}} (x>-1) y=1 when x=0




{{{(dy)/(dx)}}}{{{""=""}}}{{{(e^(4x)+1)/(e^(4x)+4x+4)}}}

{{{dy}}}{{{""=""}}}{{{expr((e^(4x)+1)/(e^(4x)+4x+4))dx}}}

{{{int(dy)}}}{{{""=""}}}{{{int(expr((e^(4x)+1)/(e^(4x)+4x+4))dx)}}}

Let u = e<sup>4x</sup> + 4x + 4
   du = (4e<sup>4x</sup> + 4)dx
   du = 4(e<sup>4x</sup> + 1)dx 
   dx = {{{(du)/(4(e^(4x) + 1))}}} 

{{{int(dy)}}}{{{""=""}}}{{{int(expr((e^(4x)+1)/(u))expr((du)/(4(e^(4x) + 1))))}}}{{{""=""}}}{{{int(expr((cross(e^(4x)+1))/(u))expr((du)/(4cross((e^(4x) + 1)))))}}}{{{""=""}}}{{{int((du)/(4u))}}}{{{""=""}}}{{{expr(1/4)int((du)/u)}}}{{{""=""}}}{{{1/4}}}ln|u| + C

y = {{{1/4}}}ln|e<sup>4x</sup> + 4x + 4| + C

And since x > -1 we can dispense with the absolute value:

y = {{{1/4}}}ln(e<sup>4x</sup> + 4x + 4) + C 

Substitute the initial condition y=1 when x=0

        1 = {{{1/4}}}ln(e<sup>4(0)</sup> + 4(0) + 4) + C
        1 = {{{1/4}}}ln(1 + 0 + 4) + C 
        1 = {{{1/4}}}ln(5) + C

1 - {{{1/4}}}ln(5) = C

Substitute for C

y = {{{1/4}}}ln(e<sup>4x</sup> + 4x + 4) + C

y = {{{1/4}}}ln(e<sup>4x</sup> + 4x + 4) + 1 - {{{1/4}}}ln(5)

You can leave it like that or 

Change the 1 to {{{4/4}}} so you can factor out {{{1/4}}}

y = {{{1/4}}}ln(e<sup>4x</sup> + 4x + 4) + {{{4/4}}} - {{{1/4}}}ln(5) 

y = {{{1/4}}}[ln(e<sup>4x</sup> + 4x + 4) + 4 - ln(5)]

Edwin</pre>