Log Integration (1 Viewer)

[DJel]

This one has me stumped;

Find the indefinite integral of 3^2x-1.

Thanks,

DJel.

 

[Forbidden.]

Image below

 

[DJel]

Thanks for that result, except the anwser is given as;

(3^2x-1)/2ln3 +C which would be (3^2x-1)/ln9 + C

 

[Forbidden.]

Thanks for that result, except the anwser is given as;

(3^2x-1)/2ln3 +C which would be (3^2x-1)/ln9 + C

You are saying the formula and the result should have been:
Image below

?

 

[DJel]

Well from the answer I would gather that the result would have to be either [(a^bx+c)/b In a] or [(a^bx+c)/In(a^b)] thats if an result for this type of question exists. In your first post were you using an actual result? I could not find it in any text book or on any website.

 

[Slidey]

This one has me stumped;

Find the indefinite integral of 3^2x-1.

Thanks,

DJel.

Changing to natural log: 3^2x = e^(ln(3^2x)) = e^(2x.ln3), thus:

Integral of: e^(2x.ln3) - 1 dx = e^(2xln3)/(2ln3) - x + C.

2ln3 can be unsimplified to ln9.

Forbidden confused his results in his initial post:

ln(a^b) = blna
ln(a*b) = lna + lnb

 

[DJel]

Sorry, I probably should have included brackets. The question is;

3^(2x-1).

DJel.

 

[watatank]

3^2x-1

= (1/3) * 3^2x

= (1/3) * 9^x

= (1/3) * e^ln9x

= (1/3) * e^x.ln9

so integrate that

I [ (1/3) * e^x.ln9 ]

= (1/3) * I [e^x.ln9]

= (1/3) * [(e^x.ln9/ln9) + c]