help with integration please =) (1 Viewer)

[danif]

heya (i put this up in the 2u forum so if you've read this twice its not you, its me)

how would you do the integration of 3^(2x-1) ?????

what's the general rule for this kind of integration?

thanks for your help

 

[*Pooja*]

is this the ans...

3^2x-1 = ?

using: y= a^x
then, dy/dx = a^x ln a

(hopefully u heard of this rule)

so in "3^2x-1" : a^x = 3^2x-1
and a = ln 3^2

therefore: 3^2x-1 = 3^2x-1 * 2 log 3
= 3^2x-1 2 log 3

(since log of x^n to the base 'b' = n log of x to the base b)

is this the ans? where did you get this question from. pm me if i got this correct. i got this topic in my half yrlys.

 

[KeypadSDM]

3^{2x - 1}
= e^{Ln[3^{2x - 1}]}
= e^{(2x - 1)Ln[3]}

Differentiation:
:. d/dx 3^{2x - 1}
= d/dx e^{(2x - 1)Ln[3]}
= 2Ln[3] * e^{(2x - 1)Ln[3]}
= 2Ln[3] * 3^{2x - 1}

Integration:

/
| 3^{2x - 1}dx
/
=
/
|e^{(2x - 1)Ln[3]}dx
/
= e^{(2x - 1)Ln[3]}/(2Ln[3]) + c
= 3^{2x - 1}/(2Ln[3]) + c