Answer
$$
\int_{0}^{1} \frac{d x}{(1+\sqrt{x})^{4}} =\frac{1}{6}
$$
Work Step by Step
$$
\int_{0}^{1} \frac{d x}{(1+\sqrt{x})^{4}}
$$
Let $u=1+\sqrt{x}$. Then $du=\frac{1}{2\sqrt{x}}dx, $ so $dx=2\sqrt{x}du$
and $dx=2(u-1)du$. When $x = 0, u = 1$; when $x = 1, u = 2$.
Thus, the Substitution Rule gives
$$
\begin{aligned}
\int_{0}^{1} \frac{d x}{(1+\sqrt{x})^{4}} &=\int_{1}^{2} \frac{1}{u^{4}} \cdot[2(u-1) d u] \\
&=2 \int_{1}^{2}\left(\frac{1}{u^{3}}-\frac{1}{u^{4}}\right) d u \\
&=2\left[-\frac{1}{2 u^{2}}+\frac{1}{3 u^{3}}\right]_{1}^{2} \\
&=2\left[\left(-\frac{1}{8}+\frac{1}{24}\right)-\left(-\frac{1}{2}+\frac{1}{3}\right)\right] \\
&=2\left(\frac{1}{12}\right)\\
&=\frac{1}{6}
\end{aligned}$$
