How do I Integrate \sqrt{1+x^4+2x^2}?

  • Thread starter Thread starter miglo
  • Start date Start date
  • Tags Tags
    Arc Integral
Click For Summary
SUMMARY

The discussion focuses on integrating the function \( \sqrt{1+x^4+2x^2} \) derived from the expression \( y=\frac{1}{3}(x^2+2)^{3/2} \). The user successfully computed the derivative \( \frac{dy}{dx}=x\sqrt{x^2+2} \) and transformed the integral into \( \int_{0}^{3}\sqrt{1+x^4+2x^2}dx \). A critical insight was recognizing that \( x^4+2x^2+1 \) can be factored as \( (x^2+1)^2 \), simplifying the integration process. The user sought assistance after struggling with the antiderivative, indicating a need for clearer substitution methods.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with derivatives and the chain rule.
  • Knowledge of algebraic manipulation and factoring polynomials.
  • Experience with integral calculus, including definite integrals.
NEXT STEPS
  • Study integration techniques involving square roots and polynomial expressions.
  • Learn about substitution methods in integral calculus.
  • Explore the use of Wolfram Alpha for solving complex integrals.
  • Practice factoring polynomials to simplify integrals.
USEFUL FOR

Students in calculus courses, educators teaching integration techniques, and anyone seeking to improve their skills in solving complex integrals.

miglo
Messages
97
Reaction score
0

Homework Statement


y=\frac{1}{3}\left(x^2+2\right)^{3/2}


Homework Equations


\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}dx


The Attempt at a Solution


\frac{dy}{dx}=x\sqrt{x^2+2}
\int_{0}^{3}\sqrt{1+\left(x\sqrt{x^2+2}\right)^{2}}dx=\int_{0}^{3}\sqrt{1+x^4+2x^2}dx
I'm stuck with that integral, not sure what the antiderivative for that function would be
Any hints at the next step would be greatly appreciated.
 
Physics news on Phys.org
Here's a hint: x4+2x2+1 = (x2+1)2
 
Wow I can't believe i didn't see that.
I tried Wolfram Alpha earlier and they used a substitution but i couldn't follow their steps so i decided to consult the Physics Forums.
Thanks Char. Limit.
 

Similar threads

Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
12
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 47 ·
2
Replies
47
Views
4K