what do I do after I've finished all 3 questions? There's no Awesome Show Points button?

This are "documents" so these ones dont have that, they are completed the moment you click on them even if you dont solve the problems. So you just gotta be happy knowing that you gave the correct answer but thats it

In question 2 is it possible to rewrite the equation to lnx * lnx instead of (lnx)^2 and integrate by parts? The problem I'm having when I try to use that method is that after I integrate lnx for the first time and substitute it back into the equation I get: ∫lnx * lnx dx = x(lnx)^2 - x - ∫(xlnx-x)/x I'm not sure how to do the new integral (∫(xlnx-x)/x).

I solved it this way. Simplify that integral by using algebra to: (∫(xlnx-x)/x) = ∫(lnx-1)dx That should make it easier to work with. I then took that and for work purposes transformed it to: ∫(lnx-1)dx = ∫(lnx-1)*1dx Once there you should be able to integrate that using another integration by parts.

In Question 1 and 3, why are they fiddling around with the order of v and u in the integration by parts equation?

Sal derived the integration by parts formula as the following: ∫(f(x)g'(x))dx = f(x)g(x) - ∫(f'(x)g(x))dx To simplify this formula, we can do a double substitution as such: ∫(f(x)g'(x))dx = f(x)g(x) - ∫(f'(x)g(x))dx u = f(x) v = g(x) du = f'(x)dx dv = g'(x)dx Rewriting the formula: ∫u dv = uv - ∫v du Hope that answered your question!

For problem 2 and 3, we never talked about compound functions. I am confused. I don't think we've learned how to do these yet.

You should be able to solve these with just integration by parts. Is there anything in particular you got stuck on?

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AP®︎/College Calculus BC

Course: AP®︎/College Calculus BC > Unit 6

Lesson 13: Using integration by parts

Integration by parts challenge

AP.CALC:

FUN‑6 (EU)

FUN‑6.E (LO)

FUN‑6.E.1 (EK)

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chang.brian710
Posted 7 years ago. Direct link to chang.brian710's post “what do I do after I've f...”
what do I do after I've finished all 3 questions? There's no Awesome Show Points button?
Button navigates to signup pageComment on chang.brian710's post “what do I do after I've f...”
(30 votes)
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- Jose Fernando Rico
  Posted 6 years ago. Direct link to Jose Fernando Rico's post “This are "documents" so t...”
  This are "documents" so these ones dont have that, they are completed the moment you click on them even if you dont solve the problems. So you just gotta be happy knowing that you gave the correct answer but thats it
  Button navigates to signup page
  (67 votes)
Grace.Ohwilleke
Posted 5 years ago. Direct link to Grace.Ohwilleke's post “In question 2 is it possi...”
In question 2 is it possible to rewrite the equation to lnx * lnx instead of (lnx)^2 and integrate by parts?

The problem I'm having when I try to use that method is that after I integrate lnx for the first time and substitute it back into the equation I get:

∫lnx * lnx dx = x(lnx)^2 - x - ∫(xlnx-x)/x

I'm not sure how to do the new integral (∫(xlnx-x)/x).
Button navigates to signup pageComment on Grace.Ohwilleke's post “In question 2 is it possi...”
(6 votes)
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- Emmet Costen
  Posted 5 years ago. Direct link to Emmet Costen's post “I solved it this way. Sim...”
  I solved it this way. Simplify that integral by using algebra to:
  
  (∫(xlnx-x)/x) = ∫(lnx-1)dx
  
  That should make it easier to work with. I then took that and for work purposes transformed it to:
  
  ∫(lnx-1)dx = ∫(lnx-1)*1dx
  
  Once there you should be able to integrate that using another integration by parts.
  Comment on Emmet Costen's post “I solved it this way. Sim...”
  (8 votes)
lovingdata
Posted 6 years ago. Direct link to lovingdata's post “In problem 2, can't we ap...”
In problem 2, can't we approach it by taking the integral of (lnx) * (lnx). So u = lnx and dv = lnx? However, when I use this approach I seem to get the wrong answer.
.............
In short, for integral( lnx * lnx ), I get lnx * (x lnx +x) - integral( lnx + 1 ), which eventually evaluates to x(lnx)^2 - 2x, but it's the wrong answer
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Sachin Shukla
Posted 6 years ago. Direct link to Sachin Shukla's post “In Question 1 and 3, why ...”
In Question 1 and 3, why are they fiddling around with the order of v and u in the integration by parts equation?
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(1 vote)
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- Aarav Shah
  Posted 4 years ago. Direct link to Aarav Shah's post “Sal derived the integrati...”
  Sal derived the integration by parts formula as the following: ∫(f(x)g'(x))dx = f(x)g(x) - ∫(f'(x)g(x))dx
  To simplify this formula, we can do a double substitution as such:
  ∫(f(x)g'(x))dx = f(x)g(x) - ∫(f'(x)g(x))dx
  u = f(x) v = g(x)
  du = f'(x)dx dv = g'(x)dx
  Rewriting the formula:
  ∫u dv = uv - ∫v du
  Hope that answered your question!
  Button navigates to signup page
  (2 votes)
devore.wes
Posted 4 years ago. Direct link to devore.wes's post “For problem 2 and 3, we n...”
For problem 2 and 3, we never talked about compound functions. I am confused. I don't think we've learned how to do these yet.
Button navigates to signup pageComment on devore.wes's post “For problem 2 and 3, we n...”
(1 vote)
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- Bryan
  Posted 3 years ago. Direct link to Bryan's post “You should be able to sol...”
  You should be able to solve these with just integration by parts. Is there anything in particular you got stuck on?
  Button navigates to signup page
  (2 votes)
Miray Atar
Posted a year ago. Direct link to Miray Atar's post “I didn't get the solution...”
I didn't get the solution to the second problem. How can we think of dv as a seperate function and where is the dx part of the antiderivative? If dv is the dx part then doesn't u supposed to be in terms of v? I couldn't get the intuition behind it and it lookks like there would be many algebraic restraints to have such solution. Can you explain it to me please?
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(1 vote)
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- cossine
  Posted a year ago. Direct link to cossine's post “So there is nothing wrong...”
  So there is nothing wrong with the solution it just a little trick.
  
  You are just the formula for integration by parts which comes from product rule.
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  (1 vote)
Hexuan Sun 8th grade
Posted 4 months ago. Direct link to Hexuan Sun 8th grade's post “On problem 2, if u = (lnx...”
On problem 2, if u = (lnx)^2, why is du = 2(lnx)/x dx
shouldn't it be only 2(lnx)/x
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(1 vote)
Answer
- obiwan kenobi
  Posted 4 months ago. Direct link to obiwan kenobi's post “du/dx is equal to 2(ln x)...”
  du/dx is equal to 2(ln x)/x. However, if you want to solve for just du, then you have to multiply both sides by dx:
  
  du/dx = 2(ln x)/x
  du = 2(ln x)/x dx
  
  All we did was that we treated du/dx like it was a ratio and then we multiplied the dx to the other side.
  
  Hope this helps!
  Comment on obiwan kenobi's post “du/dx is equal to 2(ln x)...”
  (1 vote)
Sachin Shukla
Posted 6 years ago. Direct link to Sachin Shukla's post “For the definite integrat...”
For the definite integration by parts worksheet, I was doing one that was: π∫2π of −xsin(x/2)dx. I did it correctly, but I was just wondering is there a way to do a kind of u-substitution where x = 2u so it becomes π/2∫π of −2usin(u)du? or something? I feel like it might be easier so I have to do less work to get some of the integrals.
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(1 vote)
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gesa.gq
Posted 7 years ago. Direct link to gesa.gq's post “Can you please help me wi...”
Can you please help me with this integral --- the integral of √(x+1) lnx dx
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Km Figuerrez
Posted 5 years ago. Direct link to Km Figuerrez's post “In problem 2, can't we ju...”
In problem 2, can't we just use the integral of ln(x) which is: xln(x) - x + c
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AP®︎/College Calculus BC

Course: AP®︎/College Calculus BC > Unit 6

Integration by parts challenge

Problem 1

Problem 2

Problem 3

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