How would I take the integral of this...?

∫1/9ˣ dx

= ∫9ˉˣ dx

-1/(ln9) ∫ 9ˉˣ ∙ -ln9 dx

= -1/(ln9) ∙ 9ˉˣ + C

The derivative of 9ˉˣ is 9ˉˣ ∙ln9 ∙ -1 = -ln9 ∙ 9ˉˣ, which is why you would create that exact expression inside the integral so it's the exact derivative of 9ˉˣ. When you do that, you need to also put the reciprocal in front of the integral to balance it out.

integral of 1/9^x can solved by substitution method

mltply and divide numerator and denomenator by 9^xlog(9)

we get

{9^xlog(9)}dx/{9^xlog(9).9^x}

now put t=9^x then differentiate it

dt/dx=9^xlog(9) as 9^x is in the form of a^x which has derivative

equal to a^xloga

dt=9^xlog(9)dx

now substitute dt

->{dt/(t)log(9)(t)

->dt/t^2log(9)

dt/log9t^2

now integral of 1/t^2=-1/t

now pulg in the original variables

-> integral of 1/9^x=-1/9^xlog(9) or -{9^-x/log(9)}

or u can apply formla that ∫a^x=a^x/loga

->∫9^-x=-9^-x/log9

-(9^-x/Log[9])