integration

[xXMARKXx]

Integrate

1
____________ dx
x + root(x)



I cannot figure this out. I tried "u" substitution and everything.
thanks guys

 

[mark]

you can pull out a common factor from the bottom

so it becomes

1/(x^1/2 (x^1/2 +1))

then set u = x^1/2

 

[galactus]

\(\displaystyle \L\\\int\frac{1}{x+\sqrt{x}}dx\)

Let \(\displaystyle u=\sqrt{x}, \;\ u^{2}=x, \;\ 2udu=dx\)

\(\displaystyle \L\\\int\frac{2}{(u+1)}du\)

 

[xXMARKXx]

thanks guys

 

[soroban]

Hello, xXMARKXx!

A variation of mark's approach . . .


Factor: \(\displaystyle \L\:\int\frac{dx}{x\,+\,
sqrt{x}} \;=\;\int\frac{dx}{\sqrt{x}(\sqrt{x}\,+\,1)}\;=\;\int\frac {x^{-\frac{1}{2}}dx}{\left(x^{\frac{1}{2}}\,+\,1\right)}\)

Now let \(\displaystyle u \:=\:x^{\frac{1}{2}}\,+\,1\;\;\Rightarrow\;\;du \,=\,\frac{1}{2}x^{-\frac{1}{2}}dx\;\;\Rightarrow\;\;x^{-\frac{1}{2}}dx\:=\:2\,du\)

Substitute and we have: \(\displaystyle \L\:2\int\frac{du}{u}\)