Learn to Find Derivatives and Integrals of x*lnx | Homework Help

  • Thread starter brandy
  • Start date
  • Tags
    Homework
In summary, ∫x*lnx*dx can be solved using the knowledge that the derivative of x*lnx is 1+lnx. This can be used in integration by parts to calculate the integral, which is equal to x^2 * (lnx - 1/2) / 2 + c.
  • #1
brandy
161
0

Homework Statement


find the derivative of x*lnx
and hence find
∫x*lnx*dx

The Attempt at a Solution


the derivative is 1+lnx
the integral (by parts) i worked out to be x^2 * ln(x-1/2) / 2 + c

but i don't know how you find the integral from the derivative.
help?
also is my integration by parts correct?

please and thankyou
 
Last edited:
Physics news on Phys.org
  • #2


"find the derivative of x*lnx*dx"
What is x*ln(x)*dx?

∫x*lnx*dx is a function, whose derivative is x*ln(x).
x*lnx is a function, whose derivative is 1 + ln(x).

As for your integration by parts, how on Earth did you manage to change ln(x) to ln(x - 1/2) ?
 
  • #3


both were typos
my bad

there should be no .dx for the derivative
and the integral should be x^2(lnx - 1/2) / 2 + c

oops

sorry

now will anyone answer my question?
 
  • #4


The question is asking you to use the fact that [itex]\frac{d}{dx}(x \ln x) = 1+\ln x[/itex] in calculating [itex]\int x \ln x\,dx[/itex].

How did you do your integration by parts?
 
  • #5


how do you use that fact!

i went u=lnx u'=1/x
v'=x v=x^2/2
∫x*lnx*dx=x^2lnx / 2 - ∫x^2/2x
x^2lnx / 2 - ∫x/2
x^2lnx / 2 - x^2/4
rearange
X^2 * (lnx - 1/2)*1/2
 
  • #6


If I told you that I would be telling you how to solve the problem.

Knowing f(x) and its derivative f'(x), how would you compute ∫ f(x) dx via integration by parts?
 
  • #7


but for parts u only need the derivative of lnx not x*lnx
 
  • #8


You are supposed to use the derivative of x*ln(x) to aid in the calculation of ∫ x*ln(x) dx. Yes, you can calculate this integral by other means. However, in doing so you will not be doing what was asked of you and you will get minimal points as a result.

The rules of this site (very good rules) prohibit me from giving you the solution. I can however ask some very leading questions. You have not answered my question from post #6:

Knowing f(x) and its derivative f'(x), how would you compute ∫ f(x) dx via integration by parts?
 
  • #9
Hi brandy! :smile:
brandy said:
how do you use that fact!

∫x*lnx*dx = …

Well, you obviously need to put (1 + lnx) in there somewhere, before you start. :wink:
 

1. How do I solve the integral of x*lnx*dx?

To solve this integral, we can use integration by parts. Let u = lnx and dv = x*dx. Then, du = (1/x)*dx and v = (x^2)/2. Substituting these into the integration by parts formula, we get:

∫x*lnx*dx = (x^2*lnx)/2 - ∫(x^2/2)*(1/x)*dx

Simplifying, we get:

∫x*lnx*dx = (x^2*lnx)/2 - ∫x/2*dx

Using the power rule, we can solve the second integral to get:

∫x*lnx*dx = (x^2*lnx)/2 - (x^2/4) + C

2. Do I need to use integration by parts to solve this integral?

Yes, integration by parts is the most common method for solving integrals of the form ∫x*lnx*dx. However, you can also use substitution or partial fractions in some cases.

3. Can I use a calculator to solve this integral?

Yes, some calculators have the capability to solve integrals. However, it is important to understand the steps and concepts behind solving integrals, rather than solely relying on a calculator.

4. What is the value of the constant, C, in the solution?

The value of the constant, C, represents the family of curves that satisfy the specific indefinite integral. It is often left as a variable, unless a specific initial condition is given.

5. Can you explain the concept behind integration by parts?

Integration by parts is a method for solving integrals that involves breaking down the integrand into two parts, u and dv, and using the formula ∫u*dv = uv - ∫v*du to find the solution. This method is based on the product rule for derivatives and allows us to simplify the integral into a form that is easier to solve.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
957
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
489
  • Calculus and Beyond Homework Help
Replies
19
Views
3K
  • Calculus and Beyond Homework Help
Replies
7
Views
7K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top