Question 903737
<pre>
{{{int((sec^2x*dx)/((2+tanx)(3-tanx)))}}}

[Your hint says to use t = tanx, but most teachers and books use
the letter u rather than the letter t, so I'll use u = tanx, but 
you can substitute t everywhere I have u if you like.]

let {{{u = tanx}}}, then {{{du = sec^2xdx}}}, {{{dx=du/(sec^2x)}}}

{{{int((sec^2x*(du/(sec^2x)))/((2+u)(3-u)))}}}

{{{int((cross(sec^2x)*(du/(cross(sec^2x))))/((2+u)(3-u)))}}}

{{{int(du/((2+u)(3-u)))}}}

Break {{{1/((2+u)(3-u))}}} into partial fractions:

But before doing that let's get the denominators in descending 
powers of u.
[That isn't absolutely necessary but it is customary because it 
keeps things more orderly.  Also it is customary to get the 
leading term positive, which I will do on the second term]

2+u = u+2      <--first term, just turn it around

3-u = -u+3 = -(u-3)    <--second term, factor out -1

{{{1/((u+2)(-(u-3)))}}}

{{{(-1)/((u+2)(u-3))}}}

{{{(-1)/((u+2)(u-3))}}}{{{""=""}}}{{{A/(2+u)+B/(u-3)}}}

{{{-1}}}{{{""=""}}}{{{A(u-3)+B(u+2)}}}

This has to be identically true for all u so substitute u=3
to make the first term on the right become 0

{{{-1}}}{{{""=""}}}{{{A(3-3)+B(3+2)}}}

{{{-1}}}{{{""=""}}}{{{0+5B}}}

{{{-1/5}}}{{{""=""}}}{{{B}}}

substitute u=-2 to make the second term on the right become 0:

{{{-1}}}{{{""=""}}}{{{A(-2-3))+B(-2+2)}}}

{{{-1}}}{{{""=""}}}{{{A(-5)+B(0))}}}

{{{-1}}}{{{""=""}}}{{{A(-5)+0)}}}

{{{1/5}}}{{{""=""}}}{{{A}}}

So our integral is now

{{{int(du/((2+u)(3-u)))}}} {{{""=""}}} {{{expr(1/5)*int( du/(u+2) )}}}{{{""-""}}}{{{expr(1/5)*int(du/(u-3))   }}}{{{""=""}}}

{{{expr(1/5)*ln(u+2)-expr(1/5)*ln(u-3)+C}}}{{{""=""}}}{{{expr(1/5)*(ln(u+2)^""-ln(u-3)^"")+C}}}{{{""=""}}}

Substitute {{{u = tanx}}}

{{{expr(1/5)*(ln(tanx+2)^""-ln(tanx-3)^"")+C}}}

Edwin</pre>