# Thread: Integration By Parts: LN(2x + 1)

1. ## Integration By Parts: LN(2x + 1)

Could someone show the steps of how to find the anti derivative of using the integration by parts method

ln(2x+1)

2. ## Re: Integration By Parts: LN(2x + 1)

Originally Posted by dyshanli
Could someone show the steps of how to find the anti derivative of using the integration by parts method

ln(2x+1)
See here: Integral ln(x)

4. ## Re: Integration By Parts: LN(2x + 1)

Originally Posted by dyshanli
Could someone show the steps of how to find the anti derivative of using the integration by parts method

ln(2x+1)
First you need to rewrite this as $\displaystyle \displaystyle \frac{1}{2}\int{2\ln{(2x + 1)}\,dx}$ and make the substitution $\displaystyle \displaystyle u = 2x + 1 \implies du = 2\,dx$ to turn the integral into $\displaystyle \displaystyle \frac{1}{2}\int{\ln{(u)}\,du}$, and then follow the method in ASZ's link.

5. ## Re: Integration By Parts: LN(2x + 1)

Originally Posted by Prove It
... need...
... might prefer [I would say]...

and if so, you may still find helpful to use...

... lazy integration by parts, doing without u and v.

... where (key in spoiler) ...

Spoiler:

... is the product rule for differentiation, straight continuous lines differentiating downwards with respect to x.

However, for these 'log of a linear' type integrands, note the short cut of choosing an integral of 1 that has a constant of integration: i.e. fill out the rest of the product rule shape here...

... and then the expression to subtract on the bottom row (in order to keep the lower equals sign valid) is much simpler. (As per the link above.)

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

6. ## Re: Integration By Parts: LN(2x + 1)

I still don't see the method to everyone's answers

What I've done so far is set f = ln(2x+1) df = 2/(2x+1) dg = dx g = x

With that I get f * g - Integral(df * g)

Which equals x * ln(2x+1) - Integral[ 2/(2x+1) * x]

Using the same method on Integral[ 2/(2x+1) * x] I seem to go in circles with

f = x df = 1 dg = 2/(2x+1) g = ln(2x+1)

Instead should I make f = 2/(2x+1) df = -4/(2x+1)^2 dg = x g = x^2/2

In the second case it seems like my solution is getting more complicated but in the first one it goes in circles

Is the solution to just find a way to take the Integral[ 2/(2x+1) * x ] and simpify it in someway like someone above did using the method of making

2/(2x+1) * x = 1 - 1/(2x+1)

I guess what I really want to know is should I know that when my integration by parts method isnt working then it is time to play around with the form of the term so that it becomes easier to integrate just like doing

2/(2x+1) * x = 1 - 1/(2x+1)

7. ## Re: Integration By Parts: LN(2x + 1)

Originally Posted by dyshanli
Is the solution to just find a way to take the Integral[ 2/(2x+1) * x ] and simpify it in someway like someone above did using the method of making

2/(2x+1) * x = 1 - 1/(2x+1)
Yes. (That's one good way.)

Originally Posted by dyshanli
I guess what I really want to know is should I know that when my integration by parts method isnt working then it is time to play around with the form of the term so that it becomes easier to integrate
Yeah, often.

8. ## Re: Integration By Parts: LN(2x + 1)

Originally Posted by dyshanli
I still don't see the method to everyone's answers

What I've done so far is set f = ln(2x+1) df = 2/(2x+1) dg = dx g = x

With that I get f * g - Integral(df * g)

Which equals x * ln(2x+1) - Integral[ 2/(2x+1) * x]

Using the same method on Integral[ 2/(2x+1) * x] I seem to go in circles with

f = x df = 1 dg = 2/(2x+1) g = ln(2x+1)

Instead should I make f = 2/(2x+1) df = -4/(2x+1)^2 dg = x g = x^2/2

In the second case it seems like my solution is getting more complicated but in the first one it goes in circles

Is the solution to just find a way to take the Integral[ 2/(2x+1) * x ] and simpify it in someway like someone above did using the method of making

2/(2x+1) * x = 1 - 1/(2x+1)

I guess what I really want to know is should I know that when my integration by parts method isnt working then it is time to play around with the form of the term so that it becomes easier to integrate just like doing

2/(2x+1) * x = 1 - 1/(2x+1)
It's not working because you don't integrate $\displaystyle \displaystyle \int{1 - \frac{1}{2x + 1}\,dx}$ by parts. Surely you know that the integral of $\displaystyle \displaystyle 1$ is $\displaystyle \displaystyle x$. The second term will give you a logarithm.

Of course, if you had followed my method of making a substitution in the first place, you wouldn't have this mess...

9. ## Re: Integration By Parts: LN(2x + 1)

Originally Posted by Prove It
this mess...
You're dissing Chris L T521 and Danny and Skeeter's approach (which the OP reached via my link), not mine. If you really do find it messy, then please note that my shortcut (with or without balloons) is less so.

I like your method too, though!

10. ## Re: Integration By Parts: LN(2x + 1)

Originally Posted by tom@ballooncalculus
You're dissing Chris L T521 and Danny and Skeeter's approach (which the OP reached via my link), not mine. If you really do find it messy, then please note that my shortcut (with or without balloons) is less so.

I like your method too, though!
I wouldn't necessarily call it "dissing"; he does have a valid point on how "messy" it can get (its really not that bad, but an initial substitution makes things nicer [as Prove It suggested]). In that link you posted, we just happened to help out the user where they got stuck in their work. They practically set it up correctly, and we assisted them in finishing the problem instead of introducing "something new".

11. ## Re: Integration By Parts: LN(2x + 1)

Originally Posted by Chris L T521
I wouldn't necessarily call it "dissing"; he does have a valid point on how "messy" it can get
Of course, Prove It assumed he was dissing me, but he (and Mr F?) hadn't understood the OP because he hadn't followed the link.

Originally Posted by Chris L T521
its really not that bad, but an initial substitution makes things nicer [as Prove It suggested]
Or, depending on the weather, we might think:

"a substitution is really not that bad, but going straight into parts is nicer, especially if we have the 'numerator give and take' method later on, or tom's shortcut".

Originally Posted by Chris L T521
In that link you posted, we just happened to help out the user where they got stuck in their work. They practically set it up correctly, and we assisted them in finishing the problem instead of introducing "something new".
Yes, but if you had shown a different approach as well (or several), let's hope no one would have lost their temper!

12. ## Re: Integration By Parts: LN(2x + 1)

Originally Posted by tom@ballooncalculus
Of course, Prove It assumed he was dissing me, but he (and Mr F?) hadn't understood the OP because he hadn't followed the link.

Or, depending on the weather, we might think:

"a substitution is really not that bad, but going straight into parts is nicer, especially if we have the 'numerator give and take' method later on, or tom's shortcut".

Yes, but if you had shown a different approach as well (or several), let's hope no one would have lost their temper!
My Thanks to Prove It was for pointing out that making a simple substitution at the start made the problem 'cleaner'. My contextual understanding of the word 'mess' was as a metaphor directed at the OP, rather than a critical review of the help given by other people. See Laurel and Hardy: "Well here's another nice mess you've gotten me into, Stanley".

At any rate, I think this thread can be closed.

Edit: Sheesh ..... In light of obvious continued misunderstanding, the word "mess" used in this thread is clearly a metaphor for 'still having difficulties'. 'Nuff said.