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Integration by Parts

  • Thread starter basty
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Homework Statement



Which one is correct?

##\int (3x+2) (2x+1)^{\frac{1}{2}} dx = \frac{1}{3} (3x+2)(2x+1)^{\frac{3}{2}} - \frac{1}{15} (2x+1)^{\frac{5}{2}} + C##

or

##\int (3x+2) (2x+1)^{\frac{1}{2}} dx = \frac{1}{3} (3x+2)(2x+1)^{\frac{3}{2}} - \frac{1}{5} (2x+1)^{\frac{5}{2}} + C##

?????

Homework Equations




The Attempt at a Solution



##\int (3x+2) (2x+1)^{\frac{1}{2}} dx##

Let

##u = 3x+2##

Then

##\frac{du}{dx} = 3##

or

##du = 3 \ dx##

Let

##dv = (2x+1)^{\frac{1}{2}} dx##

Then

##v = \frac{1}{3} (2x+1)^{\frac{3}{2}}##

So

##\int (3x+2) (2x+1)^{\frac{1}{2}} dx##

##= uv - \int v \ du##

##= (3x+2) \frac{1}{3} (2x+1)^{\frac{3}{2}} - \int \frac{1}{3} (2x+1)^{\frac{3}{2}} (3 \ dx)##

##= \frac{1}{3} (3x+2) (2x+1)^{\frac{3}{2}} - \int (2x+1)^{\frac{3}{2}} dx##

##= \frac{1}{3} (3x+2) (2x+1)^{\frac{3}{2}} - \frac{1}{5} (2x+1)^{\frac{5}{2}} + C##
 
Last edited:

Answers and Replies

  • #2
Tom Mattson
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You'll have an easier time with this is you don't integrate by parts. Do a u-substitution instead with u=2x+1. Much easier.

But I worked through your steps I and I got what you got.
 
Last edited:
  • #3
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You'll have an easier time with this is you don't integrate by parts.
What integration method I should use?

Do a u-substitution instead with u=2x+1. Much easier.
Do you mean changing the ##u = 2x + 1## instead of ##u = 3x + 2##?

If so, let me re-work this integration, by part.

##\int (3x + 2)(2x + 1)^{\frac{1}{2}}dx##

Let

##u = 2x + 1##

Then

##\frac{du}{dx} = 2##

Or

##du = 2 \ dx##

####

Let

##dv = (3x + 2) \ dx##

Then

##v = \frac{3}{2}x^2 + 2x##

So

##\int (3x + 2)(2x + 1)^{\frac{1}{2}}dx##

##= uv - \int v \ du##

##= (2x + 1) (\frac{3}{2}x^2 + 2x) - \int (\frac{3}{2}x^2 + 2x)(2 \ dx)##

##= (2x + 1) (\frac{3}{2}x^2 + 2x) - \int (3x^2 + 4x) dx##

##= (2x + 1) (\frac{3}{2}x^2 + 2x) - x^3 - 2x^2 + C##

##= x^3 + 4x^2 + \frac{3}{2}x^2 + 2x - x^3 - 2x^2 + C##

##= \frac{7}{2}x^2 + 2x + C##

Is it correct?

Because this is an integration by part, why use ##u = 2x + 1## instead of ##u = (2x + 1)^{\frac{1}{2}}##?

Will the result be the same or different?
 
Last edited:
  • #4
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So

##\int (3x + 2)(2x + 1)^{\frac{1}{2}}dx##

##= uv - \int v \ du##

##= (2x + 1) (\frac{3}{2}x^2 + 2x) - \int (\frac{3}{2}x^2 + 2x)(2 \ dx)##

##= (2x + 1) (\frac{3}{2}x^2 + 2x) - \int (3x^2 + 4x) dx##

##= (2x + 1) (\frac{3}{2}x^2 + 2x) - x^3 - 2x^2 + C##

##= x^3 + 4x^2 + \frac{3}{2}x^2 + 2x - x^3 - 2x^2 + C##

##= \frac{7}{2}x^2 + 2x + C##

Is it correct?
No, it's not even close. This is something that you can check for yourself. If you differentiate (7/2)x2 + 2x + C, do you get (3x + 2)(2x + 1)1/2?

When you're doing integration by parts, whatever you choose for u and dv has to multiply to give you the integrand you start with.
basty said:
Because this is an integration by part, why use ##u = 2x + 1## instead of ##u = (2x + 1)^{\frac{1}{2}}##?
You don't. I think that what Tom Mattson had in mind was an ordinary substitution.
basty said:
Will the result be the same or different?
The answer to this should be obvious.
 
  • #5
Tom Mattson
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I was just thinking u-substitution. u=2x+1, so x=(u-1)/2 and dx=du/2. Seems like less of a hassle to me because you need 2 u-substitutions in your integration by parts anyway.
 
  • #6
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∫(3x+2)(2x+1)^0.5dx
let f'(x) = (2x + 1)^0.5 ==> f(x) = ((2x + 1)^1.5)/ 3
g(x) = 2x + 2 ==> g'(x) = 2
= (2x + 2)((2x + 1)^1.5) / 3 - 2/3∫(2x + 1)^1.5 dx
= (2x + 2)((2x + 1)^1.5) / 3 - 1/3(2x + 1)^2.5 + c
= 1/3((2x + 2)(2x+1)^1.5 - (2x + 1)^ 2.5) + c
 
  • #7
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∫(3x+2)(2x+1)^0.5dx
let f'(x) = (2x + 1)^0.5 ==> f(x) = ((2x + 1)^1.5)/ 3
g(x) = 2x + 2 ==> g'(x) = 2
= (2x + 2)((2x + 1)^1.5) / 3 - 2/3∫(2x + 1)^1.5 dx
= (2x + 2)((2x + 1)^1.5) / 3 - 1/3(2x + 1)^2.5 + c
= 1/3((2x + 2)(2x+1)^1.5 - (2x + 1)^ 2.5) + c
g(x) is not (2x + 2) but (3x + 2).
 
  • #8
Ray Vickson
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Homework Statement



Which one is correct?

##\int (3x+2) (2x+1)^{\frac{1}{2}} dx = \frac{1}{3} (3x+2)(2x+1)^{\frac{3}{2}} - \frac{1}{15} (2x+1)^{\frac{5}{2}} + C##

or

##\int (3x+2) (2x+1)^{\frac{1}{2}} dx = \frac{1}{3} (3x+2)(2x+1)^{\frac{3}{2}} - \frac{1}{5} (2x+1)^{\frac{5}{2}} + C##

?????

Homework Equations




The Attempt at a Solution



##\int (3x+2) (2x+1)^{\frac{1}{2}} dx##

Let

##u = 3x+2##

Then

##\frac{du}{dx} = 3##

or

##du = 3 \ dx##

Let

##dv = (2x+1)^{\frac{1}{2}} dx##

Then

##v = \frac{1}{3} (2x+1)^{\frac{3}{2}}##

So

##\int (3x+2) (2x+1)^{\frac{1}{2}} dx##

##= uv - \int v \ du##

##= (3x+2) \frac{1}{3} (2x+1)^{\frac{3}{2}} - \int \frac{1}{3} (2x+1)^{\frac{3}{2}} (3 \ dx)##

##= \frac{1}{3} (3x+2) (2x+1)^{\frac{3}{2}} - \int (2x+1)^{\frac{3}{2}} dx##

##= \frac{1}{3} (3x+2) (2x+1)^{\frac{3}{2}} - \frac{1}{5} (2x+1)^{\frac{5}{2}} + C##
You can easily check for yourself which (if any) of the two answers is correct. Just differentiate both of them and see which (if any) gives you back the original integrand ##(3x+2)(2x+1)^{1/2}##. You should develop the habit of always doing this automatically whenever you do indefinite integrations.
 
  • #9
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4,967
You should develop the habit of always doing this automatically whenever you do indefinite integrations.
I strongly agree.
 

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