Integrating x/(sqrt[1+x^2]) - Help Needed

  • Thread starter Brewer
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In summary, the conversation is about integrating the expression x/(sqrt[1+x^2]) dx, with the answer being sqrt[1+x^2]. The person tried using the substitution u = 1+x^2, but got confused with the differential and the integral. Another person suggested that the integral becomes 0.5u^(-1/2), which the first person was unsure about how to proceed with. Eventually, it was clarified that the final step is to integrate through \int\frac{du}{2\sqrt{u}}.
  • #1
Brewer
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I have to integrate:
x/(sqrt[1+x^2]) dx

I know the answer is sqrt[1+x^2], but I can't work out how to get there. I tried the substitution u = 1+x^2, but that didn't seem to get me any where. It also looks like the differential of arsinx, but its not quite. How do I get to the answer?

Any help would be greatly appreciated.

Thank you in advance
 
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  • #2
Why didn't that subs help: it transforms it up to a scalar multiple to the integral of u^{-1/2}, which you know how to do.
 
  • #3
i managed to confuse myself, when getting du.

if u = 1 + x^2

then du/dx = 2x

so du = 2x dx yes?

so does that mean that the integral is now 0.5u^(-1/2)?

I can't remember what to do when I get to that stage.

Thanks though
 
  • #4
Brewer said:
i managed to confuse myself, when getting du.

if u = 1 + x^2

then du/dx = 2x

so du = 2x dx yes?

so does that mean that the integral is now 0.5u^(-1/2)?

I can't remember what to do when I get to that stage.

Thanks though

From what I can see, that is correct. Now simply integrate through:

[tex]\int\frac{du}{2\sqrt{u}}[/tex]

and follow through with your answer.
 
  • #5
Ta. It makes more sense now.

Thankyou, both of you
 

1. How do I integrate x/(sqrt[1+x^2])?

To integrate x/(sqrt[1+x^2]), you can use the substitution method. Let u = 1+x^2, then du = 2x dx. Rewrite the integral as ∫ (x/√u) du/2x. Simplify to ∫ √u du/2 and integrate to get (1/2)√u^3 + C. Finally, substitute back u = 1+x^2 to get the final answer as (1/2)√(1+x^2)^3 + C.

2. What is the domain of the function x/(sqrt[1+x^2])?

The function x/(sqrt[1+x^2]) is defined for all real values of x except for x = -1. This is because the denominator becomes 0 at x = -1, making the function undefined at that point. Therefore, the domain of this function is (-∞, -1) U (-1, ∞).

3. Can I use trigonometric substitution to integrate x/(sqrt[1+x^2])?

Yes, you can use trigonometric substitution to integrate x/(sqrt[1+x^2]). Let x = tanθ, then dx = sec^2θ dθ. Rewrite the integral as ∫(tanθ)/(secθ) sec^2θ dθ. Simplify to ∫sinθ dθ and integrate to get -cosθ + C. Substitute back x = tanθ to get the final answer as -cos(tan^-1(x)) + C.

4. Is there a shortcut method to integrate x/(sqrt[1+x^2])?

Yes, you can use the u-substitution method as a shortcut to integrate x/(sqrt[1+x^2]). Let u = 1+x^2, then du = 2x dx. Rewrite the integral as ∫ (x/√u) du/2x. Simplify to ∫ √u du/2 and use the formula ∫ √u du = (2/3)u^(3/2) + C. Substitute back u = 1+x^2 to get the final answer as (1/3)(1+x^2)^(3/2) + C.

5. Can I use integration by parts to solve x/(sqrt[1+x^2])?

No, integration by parts cannot be used to solve x/(sqrt[1+x^2]). This is because there is no clear choice for u and dv when using the integration by parts formula. Therefore, other integration methods such as substitution or trigonometric substitution should be used instead.

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