What is the Integral of (sqrt(x) + 1/sqrt(x))^2?

  • Thread starter Thread starter rocomath
  • Start date Start date
  • Tags Tags
    Integral
rocomath
Messages
1,752
Reaction score
1
[SOLVED] Can I do this? (integral)

\int(\sqrt{x}+\frac{1}{\sqrt{x}})^2dx sum of a product ... simplify

\int(x+2+\frac{1}{x})dx integrating ...

\frac{1}{2}x^2+2x+\ln{x}
 
Physics news on Phys.org
Looks correct to me.
 
Looks ok to me.
 
damn i love math :-] idk why i didn't think about doing it earlier.
 
Just some technicalities, \int \frac{dx}{x} = ln |x|. and \sqrt{x}^2=|x| don't think it affects anything really but if x was to be an odd power in the final answer I think it would need to be given as a parametric equation.
 
And you certainly should have "+ C"!
 
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...
Back
Top