Calculating indefinite integral

In summary: F%28u^2-2%29So first of all, let's clarify the function y: it is y = √(x+1)/√(x+2). In summary, the conversation discusses how to calculate the indefinite integral of the function y, with attempts made to use substitution methods. Multiple suggestions are made, including setting u^2=x+1 and u=\sqrt{x+2}, but ultimately the final integral remains unrecognizable and unresolved.
  • #1
Argiris
3
0

Homework Statement



hey could you help me to calculate the indefinite integral of y=√(x+1)/√(x+2)

Homework Equations





The Attempt at a Solution


tried to set x+1=u and integrate it by substitution but didnt work
 
Last edited:
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  • #2
Try to set [itex]u^2=x+1[/itex]
 
  • #3
Argiris said:

Homework Statement



hey could you help me to calculate the indefinite integral of y=√x+1/x+2

Homework Equations





The Attempt at a Solution


tried to set x+1=u and integrate it by substitution but didnt work

It is not clear what your function y is. It could be [tex] y = \sqrt{x} + \frac{1}{x} + 2, [/tex] or [tex] y = \sqrt{\displaystyle x + \frac{1}{x}} + 2, [/tex] or
[tex] y = \sqrt{\displaystyle x + \frac{1}{x} + 2}. [/tex] If you don't want to use LaTeX, you need to use brackets, so the first way I wrote above would be y = (√x) + (1/x) + 2, the second way would be y = √[x + (1/x)] + 2, and the thire way would be y = √[x + (1/x) + 2]. If I read your function using *standard* rules and priorities, it means the first way above.

RGV
 
  • #4
So now you have ∫√u/(u+1)du. Try an additional substitution.
 
  • #5
Ray Vickson said:
It is not clear what your function y is. It could be [tex] y = \sqrt{x} + \frac{1}{x} + 2, [/tex] or [tex] y = \sqrt{\displaystyle x + \frac{1}{x}} + 2, [/tex] or
[tex] y = \sqrt{\displaystyle x + \frac{1}{x} + 2}. [/tex] If you don't want to use LaTeX, you need to use brackets, so the first way I wrote above would be y = (√x) + (1/x) + 2, the second way would be y = √[x + (1/x)] + 2, and the thire way would be y = √[x + (1/x) + 2]. If I read your function using *standard* rules and priorities, it means the first way above.

RGV

I apologize for the missunderstanding the function is y=Sqrt[x+1]/Sqrt[x+2] i corrected it in the question to.
 
  • #6
micromass said:
Try to set [itex]u^2=x+1[/itex]

well i tried it and this transformed the integral into ∫2*(u^3)/√(u^2+1)du. then i set u=tanθ and the integral is transformed into 2∫(tanθ^3)*secθdθ. and couldn't take it any further..
 
  • #7
[itex] u = \sinh t [/itex] seems a better substitution.
 
  • #8
EDIT: double posting. see next post.
 
Last edited:
  • #9
Argiris said:
well i tried it and this transformed the integral into ...
I don't get that u^3.
 
  • #10
Indeed sinh does seem like a better option here
 
  • #11
Argiris said:

Homework Statement



hey could you help me to calculate the indefinite integral of y=√(x+1)/√(x+2)

Homework Equations



The Attempt at a Solution


tried to set x+1=u and integrate it by substitution but didn't work
Similar, but slightly different than what micromass suggested.

Use the substitution: [itex]u=\sqrt{x+2}\,,[/itex] then [itex]\displaystyle du=\frac{dx}{2\sqrt{x+2}}\,.[/itex]

This also gives [itex]\sqrt{x+1}=\sqrt{u^2-1}\,.[/itex]
 
  • #12

What is an indefinite integral?

An indefinite integral is a mathematical concept that represents the antiderivative of a function. It is a fundamental concept in calculus, and is used to find the original function when given its derivative.

What is the process for calculating an indefinite integral?

The process for calculating an indefinite integral involves using integration techniques, such as substitution, integration by parts, or trigonometric substitution, to find the antiderivative of a given function. This process requires knowledge of basic integration rules and formulas.

What is the difference between definite and indefinite integrals?

A definite integral has a specific interval of integration, while an indefinite integral does not. This means that a definite integral will give a specific numerical value, while an indefinite integral will give a function as the solution.

How do I know when to use which integration technique?

Choosing the appropriate integration technique depends on the form of the function. The best way to decide which technique to use is to practice and gain familiarity with each method. In general, it is helpful to look for patterns and use your knowledge of basic integration rules to determine the best technique for a given function.

Why is calculating indefinite integrals important?

Indefinite integrals are important because they allow us to find the original function when given its derivative. This is useful in many areas of science and engineering, as well as in other branches of mathematics. It also provides a way to find the area under a curve, which has many practical applications.

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