Question 877665
<pre>
{{{int(sqrt(x+4)/x,dx)}}}

Let {{{u=sqrt(x+4)}}}
Then {{{u^2=x+4}}}
and  {{{x=u^2-4}}}
and  {{{dx=2u*du}}}

Substituting in the integral:

{{{int((u/(u^2-4)),2u*du)}}}

{{{int((2u/(u^2-4)),du)}}}

{{{2*int((u^2/(u^2-4)),du)}}}

Work on {{{u^2/(u^2-4)}}} by adding -4+4 so part of the numerator
will be like the denominator:

             {{{(u^2-4+4)/(u^2-4)}}}{{{""=""}}}{{{(u^2-4)/(u^2-4)+4/(u^2-4)}}}{{{""=""}}}{{{1+4/(u^2-4)}}}

So the integral becomes:

{{{2*int((1+4/(u^2-4)),du)}}}

{{{2(int(1,du)+int(4/(u^2-4),du))}}}

{{{2(u+4*int(1/(u^2-4),du))}}}

Use the formula {{{int(1/(u^2-a^2),du)}}}{{{""=""}}}{{{expr(1/(2a))ln(abs((u-a)/(u+a)))+C}}}

{{{2(u+4*expr(1/(2*2))ln(abs((u-2)/(u+2))))+C)}}}

{{{2(u+ln(abs((u-2)/(u+2))))+C)}}}

Substitute {{{u=sqrt(x+4)}}}


{{{2(sqrt(x+4)+ln(abs((sqrt(x+4)-2)/(sqrt(x+4)+2))))+C)}}}

Edwin</pre>