since e^{ln b} = b,

b^x dx =  [ (e^{ln b}) ^x ] dx
=  e ^{(ln b) x} dx

set u = (ln b) x
then du = (ln b) dx
substitute...

=  e^u (du / ln b)
= (1 / ln b)  e^u du

solve the integral...

= (1 / ln b) ( e^u + C )
= (1 / ln b) e^u + C₂   (create new constant)

substitute back u = (ln b) x,

= ( 1 / ln b) e^{(ln b) x} + C₂
= ( 1 / ln b) ( e^{(ln b)} )^x + C₂
= ( 1 / ln b) b^x + C₂
= b^x / ln b + C₂
Q.E.D.

See also the proof of  e^u du = e^u.  PROOF

Recall that e^ln(2) = 2

2^x dx =  ( e^{ln (2)} ) ^x dx
= e^{ln (2) x} dx

set u = ln(2) x
then du = ln(2) dx
substitute:

=  e^u (du / ln 2 )
= (1 / ln 2) e^u du
= (1 / ln 2) e^u + C
substitute back...
= (1 / ln 2) e^{ln(2) x} + C
= (1 / ln 2) ( e^ln(2) )^x + C
= (1 / ln 2) 2^x + C  ANSWER

This method is actually quite fast; it just looks long because I drew it out for demostration purposes.