# Integration U-substitution

I'm having trouble understanding how they go from $$du = 3x^2$$ to $$dx = \frac{1}{3x^2} du$$.

I've set $$u$$ to be $$x^3$$ and then $$du$$ is $$3x^2 dx$$. Then $$3x^2 dx$$ becomes $$3$$. Then I get a bit lost on how they get to $$dx = \frac{1}{3x^2} du$$.

$$x^3e^{x^4}$$

• Divide both sides by $3x^2$. Feb 9 at 2:42
• I don't quite follow, which two sides? x^3 and 3x^2? Feb 9 at 2:47
• $$\require{cancel} du = 3x^2dx \ \leadsto \ \color{red}{\frac1{3x^2}}du = \color{red}{\frac1{3x^2}} 3x^2dx = \frac1{\cancel{3x^2}} \cancel{3x^2} dx = dx.$$ Feb 9 at 2:51
• If you set $u$ equal to $3x^2$, then, they are equal. If you differentiate both sides, since they were the same to begin with, they are still the same. That means that $du = 3x^2\,dx$ is an equation on the level of any other equation. So, you can divide both sides of that equation by $3x^2$, and both sides will still be equal, because you did the same thing to both of them. Feb 9 at 2:53
• @azif00 I can see how it works now, the only part i don't quite understand is why you use 1/3x^2 and it doesn't become du/3x^2? Feb 9 at 3:05

We have $$\int x^3e^{x^4}dx$$ Notice in U-Substitution method we want to eliminate all the $$x$$s and $$dx$$ instead have $$u$$s and $$du$$. so by substitution $$u=x^4$$ we achieve this purpose because:
$$u=x^4\quad\quad du=4x^3dx$$ Therefore $$x^3dx=\frac14du$$ (note that it was not necessary to extract $$dx$$ like you mentioned in your question):
$$\frac14\int e^u du=\frac14e^u+C$$ Since $$u=x^4$$ the integral is equal to $$\frac14 e^{x^4}+C$$.