\int\frac{x^3 dx}{1+\sqrt[3]{x^4+1}}=
stosujemy podstawienia
x^4+1=t^3
4x^3dx=3t^2dt
x^3dx=\frac{3}{4}t^2dt
=\frac{3}{4}\int\frac{t^2dt}{1+t}
t^2:(t+1)=t-1+\frac{1}{1+t}
-t^2-t
…
//////-t
/////////t+1
…
///////////////1
cd całki
=\int(t-1)dt+\frac{3}{4}\int\frac{1}{1+t}dt=\frac{3}{4} *\frac{t^2}{2}-\frac{3}{4}t+\frac{3}{4}ln|1+t|+C=
=\frac{3}{8} (\sqrt[3]{x^4+1})^2-\frac{3}{4}\sqrt[3]{x^4+1}+\frac{3}{4}ln|1+\sqrt[3]{x^4+1}+C