$\int {\frac{{{x^{1/2}}}}{{1 + {x^{3/4}}}}} dx$

Let $I = \int {\frac{{{x^{1/2}}}}{{1 + {x^{3/4}}}}} dx$

Let's put $x = {t^4} \Rightarrow dx = 4{t^3}dt$
therefore,$I = 4\int {\frac{{{t^2}\left( {{t^3}} \right)}}{{1 + {t^3}}}} dt = 4\int {\left( {{t^2} - \frac{{{t^2}}}{{1 + {t^3}}}} \right)} dt$

$I = 4\int {{t^2}} dt - 4\int {\frac{{{t^2}}}{{1 + {t^3}}}} dt$

$I = {I_1} - {I_2}$

${I_1} = 4\int {{t^2}} dt = 4 \cdot \frac{{{t^3}}}{3} + {C_1} = \frac{4}{3}{x^{3/4}} + {C_1}$

Now, ${I_2} = 4\int {\frac{{{t^2}}}{{1 + {t^3}}}} dt$

Again, Let's put $1 + {t^3} = z \Rightarrow 3{t^2}dt = dz$

$\Rightarrow$ ${t^2}dt = \frac{1}{3}dz = \frac{4}{3}\int {\frac{1}{z}} dz$

$= \frac{4}{3}\log |z| + {C_2} = \frac{4}{3}\log \left| {\left( {1 + {t^3}} \right)} \right| + {C_2}$

$= \frac{4}{3}\log \left| {\left( {1 + {x^{3/4}}} \right)} \right| + {C_2}$

therefore,$I = \frac{4}{3}{x^{3/4}} + {C_1} - \frac{4}{3}\log \left| {\left( {1 + {x^{3/4}}} \right)} \right| - {C_2}$

$= \frac{4}{3}{x^{3/4}} - \log \mid \left( {1 + {x^{3/4}}} \right) + C$