Evaluating Integral of $x(tan^{-1}(x))^2$

  • Context: MHB 
  • Thread starter Thread starter wonguyen1995
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
SUMMARY

The integral of \( x(\tan^{-1}(x))^2 \) requires the application of integration by parts. The suggested approach involves setting \( u = (\tan^{-1}(x))^2 \) and \( dv = x \, dx \). This method, while valid, leads to a complex solution that necessitates multiple steps, and participants in the discussion noted the absence of a straightforward result. The discussion highlights the challenges faced when attempting to evaluate this integral without clear limits of integration.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with the arctangent function and its properties.
  • Basic knowledge of calculus, particularly integral calculus.
  • Ability to manipulate and simplify expressions involving trigonometric functions.
NEXT STEPS
  • Study the method of integration by parts in detail.
  • Explore the properties and derivatives of the arctangent function.
  • Investigate alternative techniques for evaluating complex integrals, such as substitution methods.
  • Practice solving integrals involving products of polynomial and trigonometric functions.
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and integral evaluation techniques, will benefit from this discussion.

wonguyen1995
Messages
13
Reaction score
0
Find \int_{}^{} x*(tan^(-1)(x))^2\,d
 
Physics news on Phys.org
What are the limits of integration, if any?

What have you tried so far?
 
wonguyen1995 said:
Find \int_{}^{} x*(tan^(-1)(x))^2\,d

I would attempt integration by parts with $\displaystyle \begin{align*} u = \left[ \arctan{(x)} \right] ^2 \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = x\,\mathrm{d}x \end{align*}$...
 
Prove It said:
I would attempt integration by parts with $\displaystyle \begin{align*} u = \left[ \arctan{(x)} \right] ^2 \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = x\,\mathrm{d}x \end{align*}$...
I did it but it needs many steps and there is no result
 
wonguyen1995 said:
I did it but it needs many steps and there is no result

Then show us what you tried and where you got stuck...
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K