I did this exercise but I'm not sure if is done correctly. I tried to check it in Symbolab or Mathway, but didn't provide an answer. I will appreciate anyone reviewing my work and checking if is right or wrong. If it is wrong I would like to know the best way to do it. Thanks.
$$\int e^{-\sqrt[3]{t}}\;\mathrm{d}t$$
I'm doing it by substitution. I'm choosing $w=-\sqrt[3]{t}=-t^{\frac{1}{3}}$
$$w=-t^{\frac{1}{3}}$$ $$-w=t^{\frac{1}{3}}$$ $$(-w)^{3}=(t^{\frac{1}{3}})^{3}$$ $$-w^{3}=t$$
Then,
$$dw=-\frac{1}{3}t^{-\frac{2}{3}}dt$$ $$dw=-\frac{1}{3t^{\frac{2}{3}}}dt$$ $$(-3t^{\frac{2}{3}})dw=-\frac{1}{3t^{\frac{2}{3}}}dt(-3t^{\frac{2}{3}})$$ $$-3t^{\frac{2}{3}}dw=dt$$ $$-3(-w^{3})^{\frac{2}{3}}dw=dt$$ $$-3(-w)^{2}dw=dt$$ $$-3w^{2}dw=dt$$
So the integral follows,
$$\int e^{-\sqrt[3]{t}}\;\mathrm{d}t$$ $$=\int e^{w}\cdot-3w^{2}\;\mathrm{d}w$$ $$=-3\int e^{w}w^{2}\;\mathrm{d}w$$
Now, I'm doing integration by parts,
Let $u=w^2\Rightarrow du=2wdw$
Let $dv=e^{w}dw\Rightarrow v=e^{w}$
$$-3\int e^{w}w^{2}\;\mathrm{d}w$$ $$=-3[w^{2}e^{w}-\int e^{w}\cdot2w\;\mathrm{d}w]$$ $$=-3[w^{2}e^{w}-2\int e^{w}w\;\mathrm{d}w]$$ $$=-3w^{2}e^{w}+6\int e^{w}w\;\mathrm{d}w$$
Now, integration by parts again,
Let $u=w\Rightarrow du=dw$
Let $dv=e^{w}dw\Rightarrow v=e^{w}$
$$=-3w^{2}e^{w}+6\int e^{w}w\;\mathrm{d}w$$ $$=-3w^{2}e^{w}+6[we^{w}-\int e^{w}\;\mathrm{d}w]$$ $$=-3w^{2}e^{w}+6we^{w}-6\int e^{w}\;\mathrm{d}w$$ $$=-3w^{2}e^{w}+6we^{w}-6e^{w}+C$$
Remember, I choose $w=-t^{\frac{1}{3}}$. Then,
$$=-3w^{2}e^{w}+6we^{w}-6e^{w}+C$$ $$=-3e^{w}[w^{2}-2w+2]+C$$ $$=-3e^{-t^{\frac{1}{3}}}[(-t^{\frac{1}{3}})^{2}-2(-t^{\frac{1}{3}})+2]+C$$ $$=-3e^{-t^{\frac{1}{3}}}[(-1t^{\frac{1}{3}})^{2}+2t^{\frac{1}{3}}+2]+C$$ $$=-3e^{-t^{\frac{1}{3}}}[(-1)^2(t^{\frac{1}{3}})^{2}+2t^{\frac{1}{3}}+2]+C$$ $$=-3e^{-t^{\frac{1}{3}}}[1t^{\frac{2}{3}}+2t^{\frac{1}{3}}+2]+C$$ $$=-3e^{-t^{\frac{1}{3}}}[t^{\frac{2}{3}}+2t^{\frac{1}{3}}+2]+C$$ $$=-3e^{-\sqrt[3]{t}}[\sqrt[3]{t^2}+2\sqrt[3]{t}+2]+C$$