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# Evaluate the following integrals:

Question:

Evaluate the following integrals: $\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} \mathrm{dx}$

Solution:

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, $\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} d x$

$=4\left(\frac{(\sqrt[4]{x}+1)^{2}}{2}-2(\sqrt[4]{x}+1)+\log |(\sqrt[4]{x}+1)|\right)+c$