Let $\text{I}=\int\frac{1}{\sqrt{\text{x}}+\sqrt[4]{\text{x}}}\text{dx}$Let $\text{x}=\text{t}^{4}$
On differentiating both sides, we get $\text{dx}=4\text{t}^{3}\text{dt}$ $\therefore\ \text{I}=\int\frac{4\text{t}^{3}}{\sqrt{\text{t}^{4}}+\sqrt[4]{\text{t}^{4}}}\text{dt}$ $=\int\frac{4\text{t}^3}{\text{t}^{2}+\text{t}}\text{dt}$ $=4\int\frac{\text{t}^{2}}{\text{t}+1}\text{dt}$ $4\int\frac{(\text{t}-1)(\text{t}+1)+1}{\text{t}+1}\text{dt}$ $=4\int\Big[(\text{t}-1)+\frac{1}{\text{t}+1}\Big]\text{dt}$ $=4\Big[\frac{\text{t}^2}{2}-\text{t}+\log(\text{t}+1)\big]+\text{C}$ $=2\sqrt{x}-4\sqrt[4]{\text{x}}+4\log\big(\sqrt[4]{\text{x}+1}\big)+\text{C}$ Hence, $\int\frac{1}{\sqrt{\text{x}}+\sqrt[4]{\text{x}}}\text{dx}=2\sqrt{x}-4\sqrt[4]{\text{x}}+4\log\big(\sqrt[4]{\text{x}+1}\big)+\text{C}$
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