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 $dx = \frac{1}{3x^2} du$.


  • $\begingroup$ Divide both sides by $3x^2$. $\endgroup$
    – azif00
    Feb 9 at 2:42
  • $\begingroup$ I don't quite follow, which two sides? x^3 and 3x^2? $\endgroup$
    – neilmh
    Feb 9 at 2:47
  • 1
    $\begingroup$ $$\require{cancel} du = 3x^2dx \ \leadsto \ \color{red}{\frac1{3x^2}}du = \color{red}{\frac1{3x^2}} 3x^2dx = \frac1{\cancel{3x^2}} \cancel{3x^2} dx = dx.$$ $\endgroup$
    – azif00
    Feb 9 at 2:51
  • $\begingroup$ 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. $\endgroup$
    – johnnyb
    Feb 9 at 2:53
  • $\begingroup$ @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? $\endgroup$
    – neilmh
    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 $\frac14 e^{x^4}+C$.

  • $\begingroup$ Jalebe! Nice explanation Soheil. :) $\endgroup$
    – Mikasa
    Feb 9 at 7:52
  • $\begingroup$ @Miksaa mamnoon;) you're welcome! $\endgroup$ Feb 9 at 10:08

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