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).
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 $