Integration with log. (1 Viewer)

[darlking]

Hey mates.
Can you help me solve this?
Integrate 3^2x-1. I don't know how i can do this. Please show working out ?

 

[Leffife]

∫3^(2x - 1) dx

Let m = 2x - 1
i.e. dm = 2dx

Now,
= (1/2)∫[3^m].dm
= [ (3^m) / (2 ln 3) ] + C

Remember ∫3^m = (3^m) / ln 3

Now, just sub everything back in

= [ 3^(2x - 1) / ln 9 ] + C .... answer

 

[D94]

Assuming it's 3^2x-1,

Let y = 3^2x-1, then ln(y) = ln(3^2x-1) = (2x-1)ln(3) .... (by log laws)

So, y = e^(2x-1)ln(3) = e^{2ln(3)x-ln(3)}

Now, ∫e^{2ln(3)x-ln(3)} dx = 1/[2ln(3)] e^{2ln(3)x-ln(3)} + C = 3^2x-1/ln(9) + C

NB: you will need to know how to integrate an exponential function, i.e. ∫e^f(x) dx = 1/f'(x) * e^f(x) + C

 

[darlking]

I still dn't get it, can you explain it in an easier way? ):

 

[Shadowdude]

∫3^(2x - 1) dx

Let m = 2x - 1
i.e. dm = 2dx

Now,
= (1/2)∫[3^m].dm
= [ (3^m) / (2 ln 3) ] + C

Remember ∫3^m = (3^m) / ln 3

Now, just sub everything back in

= [ 3^(2x - 1) / ln 9 ] + C .... answer

Pretty sure in 2u, they don't do integration by substitution.

I still dn't get it, can you explain it in an easier way? ):

What part don't you get?

 

[D94]

I still dn't get it, can you explain it in an easier way? ):

What don't you get? Do you know your log laws? Do you know how to integrate a simpler function: e^4x wrt x? Which line of my working out don't you get? Be more specific.

 

[Focus is Key]

Pretty sure in 2u, they don't do integration by substitution.

Our teacher says you are allowed to use that method though.

 

[Focus is Key]

Our teacher says you are allowed to use Integration by Substitution. He even taught it to the 2U people and says you are allowed to use it in exams/HSC.

 

[Nws m8]

But it's hard for someone who doesn't even know the basics ....

 

[darlking]

What don't you get? Do you know your log laws? Do you know how to integrate a simpler function: e^4x wrt x? Which line of my working out don't you get? Be more specific.

Hi Hi,
Yes, I know my log laws. The second line. How did you get e? and further down, how did you change the 2ln(3)-ln(3) back to 2x-1?.... :x

 

[darlking]

But it's hard for someone who doesn't even know the basics ....

I know i Know, but not of by heart 100%, does that still count? hahahaha :x.

 

[Nws m8]

I know i Know, but not of by heart 100%, does that still count? hahahaha :x.

Lol you should know your basics of by heart mate

 

[funnytomato]

I'd assume you know:
the definition of log
log a^b = b * log a
and the integral of e^(cx+d)

 

[D94]

Hi Hi,
Yes, I know my log laws. The second line. How did you get e?

You will need to understand the relationship between the base and exponent in logarithmic and exponential functions:

(just don't confused the y's and x's in what I wrote and in the image, they are just variables)

and further down, how did you change the 2ln(3)-ln(3) back to 2x-1?.... :x

From this line:

Let y = 3^2x-1, then ln(y) = ln(3^2x-1) = (2x-1)ln(3) .... (by log laws)

So, y = e^(2x-1)ln(3) = e^{2ln(3)x-ln(3)}

Just follow the equal signs. (obviously you'll need to understand your first question before you can answer your second question)

 

[darlking]

Aww thanks for the pic! and thanks for the explanation. 10,000 HI-5's to you's all !!!

 

[Shadowdude]

You are allowed to use integration by substitution, but for 2u students it's not in their course - and technically something they're not supposed to be learning.