Question 585836
<pre>
Evaluate the indefinite integral using substitution for: 3^x.

{{{int(3^x,dx)}}}

Let u = {{{3^x}}}

ln(u) = ln({{{3^x}}})

ln(u) = x·ln(3)

Take the differential of both sides, remembering that ln(3) is a constant:

{{{(du)/u}}} = ln(3)·dx

Solve for dx by dividing both sides by ln(3)

{{{(du)/(u*ln(3))}}} = dx

Substitute u for {{{3^x}}} and {{{(du)/(u*ln(3))}}} for dx:

{{{int(3^x,dx)}}} = {{{int(u*expr((du)/(u*ln(3))))}}} = {{{int(u*du/(u*ln(3)))}}} = {{{int(cross(u)*du/(cross(u)*ln(3)))}}} = {{{int((du)/ln(3))}}} = {{{expr(1/ln(3))*int(du)}}} = {{{expr(1/ln(3))}}}u + C = {{{expr(1/ln(3))}}}u + C = {{{expr(1/ln(3))}}}{{{3^x}}} + C = {{{3^x/ln(3))}}} + C

Edwin</pre>