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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)

## Problem 1

integral, e, start superscript, x, end superscript, sine, x, d, x, equals

## Problem 2

integral, left parenthesis, natural log, x, right parenthesis, squared, d, x, equals

## Problem 3

integral, x, squared, sine, left parenthesis, pi, x, right parenthesis, d, x, equals

## Want to join the conversation?

• 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 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
• In Question 1 and 3, why are they fiddling around with the order of v and u in the integration by parts equation?
(1 vote)
• 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
• 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.
(1 vote)
• You should be able to solve these with just integration by parts. Is there anything in particular you got stuck on?
• 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?
(1 vote)
• 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.
(1 vote)
• On problem 2, if u = (lnx)^2, why is du = 2(lnx)/x dx
shouldn't it be only 2(lnx)/x
(1 vote)
• 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!
(1 vote)
• 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.
(1 vote)