How Do You Integrate x/(sqrt[1+x^2])?

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The integral of x/(sqrt[1+x^2]) dx can be solved using the substitution u = 1 + x^2. This leads to the differential du = 2x dx, allowing the integral to be rewritten as 0.5 ∫ u^(-1/2) du. The final result of the integration is sqrt[1+x^2], confirming the initial assertion. This method clarifies the process and resolves confusion regarding the substitution and integration steps.

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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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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.
 
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
 
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:

\int\frac{du}{2\sqrt{u}}

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

Thankyou, both of you
 

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