What am I doing wrong in integrating $\frac1{\sqrt{x} + \sqrt[4]{x}}$ $\displaystyle\int \dfrac{1}{\sqrt{x} + \sqrt[4]{x}}\mathrm dx$
Putting $x = t^4$ and $dx=4t^3\mathrm dt$,
$$ = \int \dfrac{4 t^3 \mathrm dt}{t^2 + 1}\\ = 4\int \dfrac{t^3 + t - t \mathrm dt}{t^2 + 1} \\= 4\int \dfrac{t^3 + t}{t^2 + 1}\mathrm dt - 4\int \dfrac{t }{t^2 + 1}\mathrm dt\\= 4\int \dfrac{t(t^2 + 1)}{t^2 + 1}\mathrm dt - 2\int \dfrac{2t }{t^2 + 1}\mathrm dt$$
$$= 4\int t\ \mathrm dt - 2\ln|t^2 + 1| + C$$
$$= 2t^2 - 2\ln|t^2 + 1| + C \\= \color{red}{2\sqrt{x} - 2\ln|\sqrt{x} + 1| + C}$$

The problem lies in the first step. After the substitution, you should have obtained$$\int\frac{4t^3}{t^2+t}\,\mathrm dt\left(=4\int t-1+\frac1{t+1}\,\mathrm dt\right).$$
