Integral containing something over the root of x²-a²

In summary, the conversation discussed how to evaluate a given integral by expressing it in different forms and using the integral table. The final solution was obtained by using a substitution and solving for the integral in two parts. A helpful website for writing LaTeX equations was also recommended.
  • #1
Telemachus
835
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
  • #2
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
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
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
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
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.
 
  • #7
Thanks Dick. BTW nice nickname :D
 
  • #8
yup, got the same answer !

35bxok1.jpg


P.S: anyone knows a program that writes LaTeX and has an intuitive interface? :)
 
  • #9
  • #10

Related to Integral containing something over the root of x²-a²

1. What is an integral containing something over the root of x²-a²?

An integral containing something over the root of x²-a² is a mathematical expression that involves finding the area under a curve defined by the function f(x) = 1/√(x²-a²). This type of integral is known as a rational function and can be solved using techniques such as substitution or integration by parts.

2. How do I solve an integral containing something over the root of x²-a²?

To solve an integral containing something over the root of x²-a², you can use techniques such as substitution or integration by parts. First, substitute u = x²-a², then find the derivative of u with respect to x. Use this to rewrite the integral in terms of u and solve as a standard integral. Alternatively, for more complex integrals, you can use integration by parts to break the integral into smaller, more manageable parts.

3. What is the purpose of solving an integral containing something over the root of x²-a²?

The purpose of solving an integral containing something over the root of x²-a² is to find the area under a curve defined by the function f(x) = 1/√(x²-a²). This can be useful in many applications, such as calculating volumes of shapes or determining the distance traveled by an object with varying velocity.

4. Are there any special cases when solving an integral containing something over the root of x²-a²?

Yes, there are special cases when solving an integral containing something over the root of x²-a². One common special case is when a = 0, which results in a simpler form of the integral. Another special case is when a is imaginary, which involves using complex numbers and can be more challenging to solve.

5. Can I use a calculator to solve an integral containing something over the root of x²-a²?

Yes, you can use a calculator to solve an integral containing something over the root of x²-a². Many scientific calculators have built-in functions for solving integrals, and there are also online integral calculators available. However, it is important to note that these calculators may not be able to solve more complex integrals and may not provide the step-by-step process for solving the integral.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
617
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
13
Views
521
  • Calculus and Beyond Homework Help
Replies
8
Views
788
  • Calculus and Beyond Homework Help
Replies
15
Views
805
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
727
  • Calculus and Beyond Homework Help
Replies
5
Views
663
  • Calculus and Beyond Homework Help
Replies
34
Views
2K
Back
Top