Here is the integral:

$\int \frac{2e^{\frac{2}{x^2}}}{x^3}dx$ or integral of $[2e^{(2/x^2)}]/x^3\,dx$

When I tried solving for it with substitution, I simply used $u=2/x^2$ whilst symbolab seemingly made $u=x$ and made $v=2/x^2$ and turned the equation into $2\;*$ integral of $[e^{(2/u)}]/2u^2\,du$ and then further manipulated the equation into $2\,* [1/2]$ integral of $- [e^v]/2\,dv$. From that step, I sort of understood how the answer $-1/2*e^{(2/x^2)}$ was found but I don't really understand how "$u$" went to "$v$" in terms of substitution. How would you solve the equation solely using "$u$" substitution? (For context: I am a high-school senior taking a brief calculus course and this is from a test that was assessing "$u$" substitution and I realized the $u/du$ pattern did not apply to this problem).


1 Answer 1


Using your substitution $u=\frac{2}{x^2}$ you should get $du=-\frac{4 \ dx}{x^3}$, $-\frac{1}{4}du=\frac{dx}{x^3}$. Also, $2e^{\frac{2}{x^2}}=2e^u$ and putting it all together, the integral becomes $-\frac{1}{2}\int e^udu $

  • $\begingroup$ this still doesn't explain how the problem becomes −1/2*∫𝑒^𝑢*𝑑𝑢 ... $\endgroup$
    – Nadia K
    May 2, 2023 at 20:03
  • $\begingroup$ @NadiaK: I've made some edit to clarify it. $\endgroup$
    – Vasili
    May 2, 2023 at 20:13

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