How to evaluate $\int x^{x^3}\cdot\left(3x^2\ln\left(x\right)+x^2\right)+2xdx$

Question

How do I evaluate $$\int x^{x^3}\cdot\left(3x^2\ln\left(x\right)+x^2\right)+2xdx$$? I know that the answer is $x^{x^3}+x^2$ (I made up this problem by taking the derivative), but how would someone (who doesn't know where it comes from) calculate this? I know that: $$\int x^{x^3}\cdot\left(3x^2\ln\left(x\right)+x^2\right)+2xdx=\int x^{x^3}\cdot\left(3x^2\ln\left(x\right)+x^2\right)dx+x^2+C$$

Answer 1

apply linearity $$\int \left(x^{x^3}(3x^2\ln x+x^2)+2x\right)dx$$$$=\frac{x^2}{2}+\int x^{x^3}\left(3x^2\ln x+x^2\right)dx$$ prepare for substitution $$=\frac{x^2}{2}+\frac{1}{3}\int (x^3)^{x^3/3}\left(\ln x^3+1\right)3x^2dx$$ substitute $u=x^3$ $$=\frac{x^2}{2}+\frac{1}{3}\int u^{u/3}\left(\ln u+1\right)du$$ prepare for substitution $$=\frac{x^2}{2}+\int e^{\frac{u}{3}\ln u}\left(\frac{\ln u+1}{3}\right)du$$ substitute $v=\frac{u\ln u}{3}$ knowing the derivative of $\frac{u\ln u}{3}$ is $\frac{1+\ln u}{2}$ $$=\frac{x^2}{2}+\int e^vdv=\frac{x^2}{2}+e^v+C$$ Undo substitutions$$=\frac{x^2}{2}+e^{x^3\ln x}+C=x^{x^3}+x^2+C$$

Answer 2

Make a $u$-substitution of $x^{x^3}$ to get: $$\int x^{x^3}\cdot\left(3x^2\ln\left(x\right)+x^2\right)+2xdx=\int 1du+x^2+C=x^{x^3}+x^2+C$$Any other answers are appreciated.