Complex Definite Integral: $\int _0^1\frac{dx}{\left(1+\sqrt{x}\right)^4}$

The substitution in the OP is equivalent to $x=u^2$. Then, $dx=2u\,du$ and we have

$$\int\frac1{(1+\sqrt x)^4}\,dx=2\int \frac{u}{(1+u)^4}\,du$$

Next, enforce the substitution $1+u=t$ to find

$$\begin{align} \int\frac1{(1+\sqrt x)^4}\,dx&=2\int \frac{u}{(1+u)^4}\,du\\\\ &=2\int \frac{t-1}{t^4}\,dt\\\\ &=\frac{2}{3t^3}-\frac{1}{t^2}+C\\\\ &=\frac{2}{3(1+u)^3}-\frac{1}{(1+u)^2}+C\\\\ &=\frac{2}{3(1+\sqrt x)^3}-\frac{1}{(1+\sqrt x)^2}+C \end{align}$$

The answer is $$\int_0^1 \frac{dx}{(1+\sqrt x\,)^4} = 1/6$$

Let $$x=(t-1)^2$$ $$dx=2(t-1)dt$$ $$\int _0^1\frac{1}{\left(1+\sqrt{x}\right)^4}dx =$$

$$\int _1^{2}\frac {2(t-1)dt}{t^4}=$$

$$2\int _1^{2} {(t^{-3}-t^{-4})dt}= \frac {1}{6}$$

Substitute $u=1+\sqrt x$ and $ du=\frac {dx} {2\sqrt x}=\frac {dx} {2(u-1)}$

$$\int _0^1\frac{1}{\left(1+\sqrt{x}\right)^4}dx=2\int _1^2\frac{\sqrt xdu} {u^4}=2\int _1^2\frac{u-1} {u^4}du=2\int_1^2\frac{du} {u^3}-2\int_1^2\frac{du} {u^4}$$