• Support PF! Buy your school textbooks, materials and every day products Here!

Integration by parts ln(2x+1)

  • Thread starter MHrtz
  • Start date
  • #1
53
0

Homework Statement


Take the integral of the following:

1. ln(2x+1)
2. arctan4x
3. ecosxsin2x evaluated from 0 to pi


The Attempt at a Solution



1. took the derivative of ln(2x+1) and integrated dx. my solution was:

xln(2x +1) + x + [(2x + 1)-2]/2 + C

The books answer was:

(1/2)(2x + 1)ln(2x + 1) - x + C

2. took the derivative of arctan4x and the integrated dx. My solution was:

xarctan4x - (1+ x2)-2 + C

The books answer was:

xarctan 4x - (1/8)ln(1 + 16x2) + C

3. I derived sin2x and integrated ecosx but I got something more complex than what I started with.
 

Answers and Replies

  • #2
33,631
5,288

Homework Statement


Take the integral of the following:

1. ln(2x+1)
2. arctan4x
3. ecosxsin2x evaluated from 0 to pi


The Attempt at a Solution



1. took the derivative of ln(2x+1) and integrated dx. my solution was:

xln(2x +1) + x + [(2x + 1)-2]/2 + C
I'm not sure that you understand how integration by parts works. When you do this, you get a product of functions (xln(2x + 1) in this case) and another integral, that should be simpler than the one you started with.

[tex]\int \ln{(2x+1)}dx = x~ln(2x + 1) - \int \frac{2xdx}{2x + 1}[/tex]
You didn't mention getting the integral on the right. If you did get it, I don't think you integrated correctly.
The books answer was:

(1/2)(2x + 1)ln(2x + 1) - x + C

2. took the derivative of arctan4x and the integrated dx. My solution was:

xarctan4x - (1+ x2)-2 + C

The books answer was:

xarctan 4x - (1/8)ln(1 + 16x2) + C

3. I derived sin2x and integrated ecosx but I got something more complex than what I started with.
 
  • #3
53
0
I did get the integral on the right then I made u = 2x + 1, du = 2 dx, and (u - 1) = 2x
 
  • #4
Char. Limit
Gold Member
1,204
14
For the third integral, start by changing sin(2x) to 2 sin(x) cos(x) using the double angle rule, then substitute u=cos(x). Only then should you integrate by parts.
 
  • #5
53
0
If u = cosx then does this mean that dv = 2sinx ecosx dx?
 
  • #6
Char. Limit
Gold Member
1,204
14
No, I mean make a u-substitution first. THEN integrate by parts. Use s=cos(x) if you're too used to using u and dv for integration by parts.
 
  • #7
33,631
5,288
I did get the integral on the right then I made u = 2x + 1, du = 2 dx, and (u - 1) = 2x
So it looks like you are doing an ordinary substitution. There are two other approaches that are more direct.
1. Rewrite the integrand of the integral on the right like so.
[tex]\frac{2x}{2x + 1} = \frac{2x + 1 - 1}{2x + 1} = \frac{2x + 1}{2x + 1} - \frac{1}{2x + 1} = 1 - \frac{1}{2x + 1}[/tex]
So
[tex]-\int \frac{2xdx}{2x+1} = -\int\left(1 - \frac{1}{2x + 1} \right)dx[/tex]
This is an easy integral.
2. Use polynomial division to divide 2x by 2x + 1. This results in 1 - 1/(2x + 1), the same as above.
 
  • #8
33,631
5,288
For the third integral, start by changing sin(2x) to 2 sin(x) cos(x) using the double angle rule, then substitute u=cos(x). Only then should you integrate by parts.
If u = cosx then does this mean that dv = 2sinx ecosx dx?
Yes. When you integrate dv you will get -ecosx. Integration by parts results in the following:
[tex]-cosx~e^{cosx} - \int e^{cosx}~sinx~dx[/tex]

The integral on the right can be done with a substitution.
 
  • #9
Char. Limit
Gold Member
1,204
14
Oh right... I guess you could do integration by parts that way too.
 
  • #10
53
0
I tried your method for the third integral but still got a wrong answer. I used a calculator and determined that the actual answer was e - (1/e). Here's what I did:

s = cosx
ds = -sinx dx

[itex]\int[/itex]2sinxcosxecosxdx = -2[itex]\int[/itex]sesds

u = s
du = ds
dv = esds
v = es

-2ses - [itex]\int[/itex]esds

-2cosxecosx + [itex]\int[/itex]sinxecosx dx

Let me know if that much is right so I can continue.
 
  • #11
33,631
5,288
I tried your method for the third integral but still got a wrong answer. I used a calculator and determined that the actual answer was e - (1/e). Here's what I did:

s = cosx
ds = -sinx dx

[itex]\int[/itex]2sinxcosxecosxdx = -2[itex]\int[/itex]sesds

u = s
du = ds
dv = esds
v = es

-2ses - [itex]\int[/itex]esds

-2cosxecosx + [itex]\int[/itex]sinxecosx dx
In the line above, the + in the middle should be =. Also, there should be a coefficient of 2 on the integral at the right. These two errors would cause you to get an incorrect answer.
Let me know if that much is right so I can continue.
 
  • #12
53
0
-2cosxecosx - 2[itex]\int[/itex]sinxecosx dx

u = cosx
du = -sinx dx

2[itex]\int[/itex]eudu

-2cosxecosx + 2eu

[-2cosxecosx + 2ecosx] evaulated from 0 to pi

What about this?
 
  • #13
33,631
5,288
I get 4/e, which is the same as what wolframalpha gets. The value you reported earlier (e - 1/e) is apparently incorrect.
 
  • #14
53
0
Ok. So I get 4/e now as my final answer. My mistake was that I dropped a negative while carrying out my evaluation. For the first integral I tried a different attempt but the answer is still off:

u = ln(2x+1)
du = 2/(2x+1) dx
dv = dx
v = x

xln(2x+1) - [itex]\int[/itex] 2x/(2x+1) dx

u = 2x + 1
du = 2 dx
x = (u-1)

xln(2x+1) - (1/2)[itex]\int[/itex](u-1)/u du

xln(2x+1) - (1/2)[itex]\int[/itex]1 - u-1 du

xln(2x+1) - (1/2)x - (1/2)ln(2x + 1) + C

Can this be simplified to look like the book's answer or am I still integrating incorrectly?
 
Last edited:
  • #15
33,631
5,288
Ok. So I get 4/e now as my final answer. My mistake was that I dropped a negative while carrying out my evaluation. For the first integral I tried a different attempt but the answer is still off:

u = ln(2x+1)
du = 2/(2x+1) dx
dv = dx
v = x

xln(2x+1) - [itex]\int[/itex] 2x/(2x+1) dx

u = 2x + 1
du = 2 dx
x = (u-1)
The line above is wrong. u = 2x + 1 ==> x = (u - 1)/2
xln(2x+1) - (1/2)[itex]\int[/itex](u-1)/u du

xln(2x+1) - (1/2)[itex]\int[/itex]1 - u-1 du

xln(2x+1) - (1/2)x - (1/2)ln(2x + 1) + C

Can this be simplified to look like the book's answer or am I still integrating incorrectly?
No and yes. See above.
 
  • #16
Char. Limit
Gold Member
1,204
14
Ok. So I get 4/e now as my final answer. My mistake was that I dropped a negative while carrying out my evaluation. For the first integral I tried a different attempt but the answer is still off:

u = ln(2x+1)
du = 2/(2x+1) dx
dv = dx
v = x

xln(2x+1) - [itex]\int[/itex] 2x/(2x+1) dx

u = 2x + 1
du = 2 dx
x = (u-1)
It's actually 2x = (u - 1). Which still gives the correct integral in the next line, I believe.

[itex]x ln(2x+1) - \frac{1}{2}\int\frac{u-1}{u} du[/itex]

xln(2x+1) - (1/2)[itex]\int[/itex]1 - u-1 du

xln(2x+1) - (1/2)x - (1/2)ln(2x + 1) + C

Can this be simplified to look like the book's answer or am I still integrating incorrectly?
 

Related Threads on Integration by parts ln(2x+1)

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
10
Views
5K
Replies
4
Views
995
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
9
Views
11K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
1
Views
9K
  • Last Post
Replies
19
Views
1K
Replies
5
Views
2K
Top