Learn How to Integrate xlnx Using the Integration by Parts Method

In summary, the conversation involves a request for help with integrating xlnx and using integration by parts. The attempt at a solution includes letting u = x and dv = ln x, but the integral of lnx is not known. Another suggestion is made to let u = lnx and dv = xdx instead. The final attempt involves integrating from 1 to 2 and correctly evaluates to 2ln2 - 3/4.
  • #1
ricekrispie
22
0
little help? simple question...

Homework Statement




Integrate:
xlnx dx

use integration by parts


Homework Equations





The Attempt at a Solution



let u = x du = dx
dv = ln x v = ??
 
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  • #2
If you don't know the integral of lnx...try letting u=lnx and dx=x dx instead of the way you wrote it
 
  • #3

Homework Statement




Integrate:
xlnx dx



Homework Equations





The Attempt at a Solution



let u = x du = dx
dv = ln x v = ??
 
  • #4
How about

Let u = lnx
du = 1/x
dv = xdx
v = x^2 / 2
 
  • #5
if I have to integrate it from 1 to 2 where are these plugged in?
 
  • #6
[tex]\int_a^b udv = uv|_a^b - \int_a^b vdu[/tex]
Or in other words
[tex]\int_a^b udv = u(b)v(b) - u(a)v(a) - \int_a^b vdu[/tex]
 
  • #7
my attempt

i wonder if this worked...

[tex]\int_1^2 x lnx dx = lnx (x^2/2)|_1^2 - \int_1^2 (x^2/2)* (1/x) dx [/tex]
[tex] = 2ln2 - ((1/4)(x^2))|_1^2 [/tex]

= 2ln2 - (5/4)
??
 
  • #8
ricekrispie said:
i wonder if this worked...

[tex]\int_1^2 x lnx dx = lnx (x^2/2)|_1^2 - \int_1^2 (x^2/2)* (1/x) dx [/tex]
[tex] = 2ln2 - ((1/4)(x^2))|_1^2 [/tex]

= 2ln2 - (5/4)
??
Almost! [itex](1/4)(x^2)[/itex] when x= 2 is 1. when x= 1, it is 1/4 so the last part is -(1- 1/4)= -3/4, not -5/4.
 
  • #9
check your evaluation for the last part

it might be me but you did the integration correctly, gj.
 
  • #10
Almost...

[tex]x^2|^2_1[/tex] Is 3 not 5.
 
  • #11
hehe i noticed a lack of parenthesis... thanks for helping :smile:
 

1. What is integration by parts?

Integration by parts is a method for evaluating integrals that involve products of functions. It is based on the product rule from differentiation and involves breaking down the integral into smaller, more manageable parts.

2. Why is it important to learn how to integrate using the integration by parts method?

The integration by parts method is important because it allows us to solve a wide variety of integrals that cannot be solved using other methods. It also helps us to evaluate integrals more efficiently and accurately.

3. How do you know when to use integration by parts?

You should use integration by parts when the integral involves the product of two functions and one of the functions becomes simpler after differentiating and one becomes simpler after integrating. This is known as the LIATE rule.

4. What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are functions and dv and du are their respective differentials. This formula is derived from the product rule for differentiation.

5. Can you provide an example of integrating using the integration by parts method?

Sure, an example of integrating using the integration by parts method would be ∫xlnx dx. Using the LIATE rule, we let u = ln x and dv = x dx. This gives us du = 1/x and v = x^2/2. Plugging these values into the formula, we get ∫xlnx dx = x^2/2 * ln x - ∫x^2/2 * 1/x dx. Simplifying this further, we get ∫xlnx dx = x^2/2 * ln x - ∫x/2 dx. By evaluating the integrals, we get the final answer of x^2/2 * ln x - x^2/4 + C.

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