Integral containing something over the root of x²-a²

  • Thread starter Telemachus
  • Start date
  • #1
832
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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619
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.
 
  • #3
hunt_mat
Homework Helper
1,742
26
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
 
  • #4
832
30
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!
 
  • #5
832
30
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]
 
  • #6
Dick
Science Advisor
Homework Helper
26,260
619
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.
 
  • #7
832
30
Thanks Dick. BTW nice nickname :D
 
  • #8
275
0
yup, got the same answer !

35bxok1.jpg


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

Related Threads on Integral containing something over the root of x²-a²

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
10
Views
6K
  • Last Post
Replies
1
Views
22K
  • Last Post
Replies
14
Views
2K
Replies
6
Views
10K
Replies
25
Views
3K
Replies
5
Views
2K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
912
Top