Integrate using $t$ substitution integrate this $\int \frac{1}{\sqrt{x+1}+\sqrt{x}}$
I took $\sqrt{x} = t$
Differentiated it
So $\frac{1}{2\sqrt{x}}=\frac{dt}{dx}$
Then I don't know how to proceed
I did some steps and it got too complicated
I also took  $\frac{1}{\sqrt{x+1}+\sqrt{x}} = t$
But that also got complicated
A hint would be perfect

A COMPLICATED SOLUTION:
Let $$\sqrt{1+x}+\sqrt{x}=u, \quad \sqrt{1+x}-\sqrt{x}=1/u \implies 2\sqrt{x}=u-1/u~~~~(*)$$
So we get $dx=\frac{(u^4-1)}{4u^2} du$
So $$I=\int \frac{dx}{\sqrt{1+x}+\sqrt{x}}=\frac{1}{2}\int (1-\frac{1}{u^4}) dx$$
$$\implies I= \frac{1}{2} \left(u+\frac{1}{3u^3}\right) =\frac{1}{2}\left(\sqrt{1+x}+\sqrt{x}+\frac{1}{3}(\sqrt{1+x}-\sqrt{x})^3\right)$$
$$\implies I=\frac{2}{3}[(1+x)^{3/2}-x^{3/2}].$$
Such tranformation as (*) solve more interesting problems.
