Fun problem: ? x / (x^2 + 6x + 10) dx

  • Context: Undergrad 
  • Thread starter Thread starter Matt Jacques
  • Start date Start date
  • Tags Tags
    Dx Fun
Click For Summary
SUMMARY

The integral of the function x / (x^2 + 6x + 10) can be solved using integration by parts and substitution methods. The final result is expressed as 1/2(ln[10 + x(6 + x)] - 6 arctan[3 + x]). The denominator x^2 + 6x + 10 is irreducible over the reals and can be transformed using the substitution u = x + 3, simplifying the integration process. This approach highlights the utility of both integration by parts and partial fractions in solving complex integrals.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with substitution methods in calculus.
  • Knowledge of logarithmic and arctangent functions.
  • Concept of irreducible polynomials in calculus.
NEXT STEPS
  • Study advanced integration techniques, focusing on integration by parts.
  • Learn about partial fraction decomposition in calculus.
  • Explore the properties of irreducible polynomials and their implications in integration.
  • Practice solving integrals involving logarithmic and trigonometric functions.
USEFUL FOR

Students and educators in calculus, mathematicians interested in integration techniques, and anyone looking to enhance their problem-solving skills in advanced mathematics.

Matt Jacques
Messages
81
Reaction score
0
Fun problem: ? x / (x^2 + 6x + 10) dx

Integration by parts proves 1=1! My mathematical fame is at hand! So how would you do this one?
 
Physics news on Phys.org
\int\frac{x}{x^2 + 6x + 10}{\rm d}x = \frac12\left(\ln[10 + x(6 + x)]-6 \arctan [3 + x] \right)


So, what does this have to do with 1=1 (which is selfevidently true anyway)?
 
Last edited:
Integration by parts proves 1=1? In other words, you used integration by parts twice, the second time reversing your choice for u and dv so the two cancelled!

"Partial fractions" is what you need here. The denominator, x^2 + 6x + 10, is "irreducible" over the real numbers. It is the same as
x^2+ 6x+ 9+ 1= (x+3)^2+ 1. I would recommend the substitution
u= x+ 3 so that du= dx, x= u- 3 and the problem becomes integrating
(u-3)/(u^2+1)= u/(u^2+1)- 3/(u^2+1).

The first of those can be done by the further substitution v= u^2+1 and the second is a simple arctangent.
 
OOops, I forgot that x could be expressed in terms of u.
 
Last edited:

Similar threads

Undergrad Ambiguity of the term "indefinite integral"
  • · Replies 11 ·
Replies
11
Views
1K
Undergrad Derive the orthonormality condition for Legendre polynomials
  • · Replies 30 ·
2
Replies
30
Views
1K
Graduate Non solvable integral? (dx/dt)^2 dt
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad Understanding Reduction Formula
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad PF Integral Bee: Share Interesting/Quirky Integrals!
  • · Replies 1 ·
Replies
1
Views
1K
High School Integration by parts of inverse sine, a solved exercise, some doubts...
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Integration with different infinitesimal intervals
  • · Replies 31 ·
2
Replies
31
Views
4K
Undergrad How Do You Integrate 1/√(x^3 + x^2 + x + 1) dx?
  • · Replies 8 ·
Replies
8
Views
2K
High School Integration by Parts, an introduction I get confused with
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Substitution in a definite integral
  • · Replies 6 ·
Replies
6
Views
3K