Integral containing something over the root of x²-a²

  • Thread starter Thread starter Telemachus
  • Start date Start date
  • Tags Tags
    Integral Root
Click For Summary

Homework Help Overview

The discussion revolves around evaluating the integral \(\int \frac{x+3}{\sqrt{x^2+2x}}dx\) and expressing it in forms involving square roots of quadratic expressions. Participants are exploring methods of integration and substitution in the context of calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to complete the square and substitute variables to simplify the integral. Some participants suggest splitting the integral into simpler parts or using hyperbolic substitutions. Others express confusion about the complexity of the integration process.

Discussion Status

Participants have shared various approaches to the integral, with some expressing satisfaction with their solutions while others are still exploring different methods. There is acknowledgment of helpful guidance regarding substitutions and integral splitting.

Contextual Notes

Some participants note the challenge of the integral and the potential for different interpretations of the problem setup. There is also mention of preferences in notation and expression forms.

Telemachus
Messages
820
Reaction score
30

Homework Statement


Evaluate the next integrals, expressing it previously in the forms that contains [tex]\sqrt[ ]{a^2+u^2}[/tex], [tex]\sqrt[ ]{a^2-u^2}[/tex] and [tex]\sqrt[ ]{u^2-a^2}[/tex], and then solve it using the integral table;

The exercise which I couldn't solve:

[tex]\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sqrt[ ]{x^2+2x}}dx[/tex]

The Attempt at a Solution


I started by completing the square:

[tex]x^2+2x=(x+1)^2-1[/tex]

Then

[tex]u^2=(x+1)^2[/tex]
[tex]u=(x+1)\Rightarrow{x+3=u+2}[/tex]
[tex]du=dx[/tex]

[tex]\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sqrt[ ]{x^2+2x}}dx=\displaystyle\int_{}^{}\displaystyle\frac{u+2}{\sqrt[ ]{u^2-1}}du[/tex]

From here I've tried to solve it by parts
[tex]t=u+2[/tex]
[tex]dt=du[/tex]

[tex]dv=\displaystyle\frac{1}{\sqrt[ ]{u^2-1}}du[/tex]
[tex]v=\ln|u+\sqrt[ ]{u^2-1}|[/tex]

[tex]\displaystyle\int_{}^{}\displaystyle\frac{u+2}{\sqrt[ ]{u^2-1}}du=(u+2)\ln|u+\sqrt[ ]{u^2-1}|-\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du[/tex]

If I try again by parts with [tex]-\displaystyle\int_{}^{}\ln|u+\sqrt[ ]{u^2-1}|du[/tex] it gets more complicated.

So, what do you say?

Bye there.
 
Physics news on Phys.org
You can do integral u/sqrt(u^2-1) with a simple substitution. So split (u+2)/sqrt(u^2-1) up into two integrals.
 
The integral
[tex] \int\frac{u+2}{\sqrt{u^{2}-1}}du[/tex]
Can be computed by the substitution
[tex] u=\cosh v[/tex]
Everything should come out in the wash then
 
Thank you both.

This is what I wasn't seeing:
You can do integral u/sqrt(u^2-1) with a simple substitution. So split (u+2)/sqrt(u^2-1) up into two integrals.

Bye there!
 
I've arrived to this solution:

[tex]\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sqrt[ ]{x^2+2x}}dx=\sqrt[ ]{(x+1)^2-1}+2\ln|x+1+\sqrt[ ]{(x+1)^2-1}|+C[/tex]
 
Telemachus said:
I've arrived to this solution:

[tex]\displaystyle\int_{}^{}\displaystyle\frac{x+3}{\sqrt[ ]{x^2+2x}}dx=\sqrt[ ]{(x+1)^2-1}+2\ln|x+1+\sqrt[ ]{(x+1)^2-1}|+C[/tex]

Seems ok to me. I would write x^2+2*x instead of (x+1)^2-1. But that's up to you.
 
Thanks Dick. BTW nice nickname :D
 
yup, got the same answer !

35bxok1.jpg


P.S: anyone knows a program that writes LaTeX and has an intuitive interface? :)
 

Similar threads

Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
4
Views
3K
Replies
6
Views
3K