How can integration by parts be used to solve this integral?

In summary, to solve the integral of x^2ln(x)dx, we use integration by parts with u=ln(x), du=1/x, dv=x^2dx, and v=x^3/3. This gives us the solution x^3/3ln(x) - (1/3)\int x^2dx, which simplifies to x^3/3ln(x) - (1/3)(x^3/3) + C.
  • #1
jpd5184
76
0

Homework Statement



integral of x^2ln(x)dx

Homework Equations





The Attempt at a Solution



u=ln(x)
du= 1/x
dv=x2dx
x^3/3

integral x^2ln(x)dx = ln(x)x^3/3-intergral(x^3/3)(1/x)
 
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  • #2
jpd5184 said:

Homework Statement



integral of x^2ln(x)dx

Homework Equations





The Attempt at a Solution



u=ln(x)
du= 1/x
dv=x2dx
x^3/3

integral x^2ln(x)dx = ln(x)x^3/3-intergral(x^3/3)(1/x)
You're on the right path. Just continue what you're doing, but simplify the integrand on the right. Don't forget dx or the constant of integration, though.

Here's the work in LaTeX. Click the equation to see what I did.
[tex]\int x^2 ln(x)dx = \frac{x^3}{3}ln(x) - \frac{1}{3}\int \frac{x^3}{x} dx + C[/tex]
 
  • #3
The natural logarithm is also a function recognized by LaTeX. It has the code [itex] \ln x [/itex]. Its inverse, the exponential function in the base e also has a code [itex] \exp x [/itex].
 

FAQ: How can integration by parts be used to solve this integral?

What is "Integration by parts"?

"Integration by parts" is a common method used in calculus to solve integrals. It involves breaking down a complicated integral into simpler parts and then applying the integration formula to each part.

When should "Integration by parts" be used?

"Integration by parts" should be used when the integral involves a product of two functions, one of which can be easily integrated and the other can be differentiated. This method is especially useful when the integral cannot be solved by other methods such as substitution or partial fractions.

How do you apply "Integration by parts"?

To apply "Integration by parts", you need to identify the two functions in the integral - one that can be easily integrated and one that can be differentiated. Then, use the formula: ∫u dv = uv - ∫v du, where u is the function to be differentiated and dv is the function to be integrated. Repeat this process until the integral can be solved.

Are there any limitations to "Integration by parts"?

Yes, there are some limitations to "Integration by parts". This method can only be used when the integral involves a product of two functions. It is also not effective when the integral has a complicated form or when the derivative of the function to be differentiated does not have a pattern.

Can "Integration by parts" be used for definite integrals?

Yes, "Integration by parts" can be used for definite integrals. After applying the integration formula, you can evaluate the definite integral using the limits of integration. However, it is important to be careful with the limits of integration when using this method.

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