Evaluate the following integrals: $\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} \mathrm{dx}$
Let $I=\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} d x$
$\Rightarrow I=\int \frac{1}{\sqrt[4]{x}(\sqrt[4]{x}+1)} d x$
Multiplying and dividing by $\sqrt{x}$
$\Rightarrow I=\int \frac{x^{\frac{1}{2}}}{x^{\frac{3}{4}}(\sqrt[4]{x}+1)} d x$
Let $\sqrt[4]{\mathrm{x}}+1=\mathrm{t} \Rightarrow \frac{1}{4} \mathrm{x}^{-\frac{3}{4}} \mathrm{dx}=\mathrm{dt}$
So, $\Rightarrow I=4 \int \frac{(t-1)^{2}}{t} d t$
$\Rightarrow \mathrm{I}=4 \int \frac{\mathrm{t}^{2}-2 \mathrm{t}+1}{\mathrm{t}} \mathrm{dt}$
$\Rightarrow \mathrm{I}=4 \int\left(\mathrm{t}-2+\frac{1}{\mathrm{t}}\right) \mathrm{dt}$
$\Rightarrow \mathrm{I}=4\left(\frac{\mathrm{t}^{2}}{2}-2 \mathrm{t}+\log |\mathrm{t}|\right)+\mathrm{c}$
$\Rightarrow I=4\left(\frac{(\sqrt[4]{x}+1)^{2}}{2}-2(\sqrt[4]{x}+1)+\log |(\sqrt[4]{x}+1)|\right)+c$
Therefore, $\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} d x$
$=4\left(\frac{(\sqrt[4]{x}+1)^{2}}{2}-2(\sqrt[4]{x}+1)+\log |(\sqrt[4]{x}+1)|\right)+c$