Integrals: Evaluating ∫ x^{m}* (ln(x))^{2} using Integration by Parts

  • Thread starter Thread starter Justabeginner
  • Start date Start date
  • Tags Tags
    Variables
Click For Summary
SUMMARY

The integral ∫ x^{m} * (ln(x))^{2} can be evaluated using integration by parts. The correct approach involves setting u = (ln(x))^{2} and v = x^{m+1}/(m+1). The resulting expression simplifies to ∫ x^{m} * (ln(x))^{2} dx = (ln(x))^{2} * (x^{m+1})/(m+1) - (2/(m+1)) * ∫ x^{m} * ln(x) dx. A second application of integration by parts is necessary to resolve the remaining integral.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts
  • Familiarity with logarithmic functions and their properties
  • Knowledge of polynomial functions and their integration
  • Ability to manipulate integrals involving logarithmic expressions
NEXT STEPS
  • Practice additional examples of integration by parts with logarithmic functions
  • Learn how to evaluate integrals involving products of polynomials and logarithms
  • Study the reduction formula for integrals of the form ∫ x^{n} * ln(x) dx
  • Explore advanced integration techniques, including multiple applications of integration by parts
USEFUL FOR

Students studying calculus, particularly those focusing on integral calculus and integration techniques. This discussion is beneficial for anyone looking to master the evaluation of complex integrals involving logarithmic functions.

Justabeginner
Messages
309
Reaction score
1

Homework Statement


Evaluate the integral: ∫ x^{m}* (ln(x))^{2} It's said as ln squared x. Sorry if I miswrote it.

Homework Equations


∫udv= uv - ∫vdu

The Attempt at a Solution


u= (ln(x)^2)
v= x^{m+1}/(m+1)
du= 2lnx/(x)
dv= x^{m} * dx

- ∫2lnx * (x^{m+1})/(x*(m+1)) + [(ln(x)^{2}) (x^{m+1})/(m+1)}
-(x^{m+1})/(m+1) * ∫ 2lnx/(x) + (ln(x)^{2})

I am stuck here. I feel like I'm not doing this right, and I'm sure I'm not. Can I get some guidance as to if I'm even doing this right? Thank you so much.
 
Physics news on Phys.org
Up to here, you're okay:
$$\int x^m (\ln x)^2\,dx = (\ln x)^2\frac{x^{m+1}}{m+1} - \frac{2}{m+1} \int x^m\ln x\,dx.$$ Your next step is wrong. You can't pull anything that depends on ##x## out of the integral.

Did you notice you've ended up with essentially the same integral as before except the exponent of the log has gone done by one? This suggests you should try integrating by parts one more time.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
4K
  • · Replies 54 ·
2
Replies
54
Views
16K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 5 ·
Replies
5
Views
1K