What is the Integral of 1 / 1 + sqrt [x] Using the Chain Rule and Substitution?

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Hi, I am a newbie here and would like to ask you stg. integral of 1 / 1 + sqrt [x]

I used chain rule and used x = u^2 is that true?

Thanx for any replies
 
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Do the substitutions that neutrino and morphism suggested.
 


Can't you use integration by parts?
 


I'm doing this in my head and not fully awake, so I could be wrong.
we have 1/(1+sqrt(x)) You can use the substitution that you've shown.
u^2=x
2udu=dx
So we end up with 2u/1+u
remember you can substitute more than once...
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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