Definite integral substitution problem

Question

I had to solve the following definite integral, i solved it by substitution but I checked many times and I can't find out why the result I get is wrong. I know it can be solved by parts, but I want to understand what is wrong with my reasoning specifically. The integral is the following: \[ I = \int_2^4 \frac{\arctan \frac{2}{x}}{x^3} dx \] I thought about substituting \[ t = \frac{2}{x} \qquad \implies \qquad dt = -\frac{2}{x^2} dx \] Then I rewrote the integral as follows \[ I = -\frac{1}{2} \int_2^4 \frac{\arctan \frac2x}{x} \cdot \left(-\frac{2}{x^2}\right) dx \] If $x=2$ then $t=\frac{2}{2}=1$, Whilst if $x=4$ then $t=\frac{2}{4}=\frac{1}{2}$; also by the definition of $t$ I get that $x=\frac{2}{t}$ so by making the substitution I get \[ I =-\frac{1}{2} \int_{1}^{\frac 12} \frac{\arctan t}{\frac{2}{t}} dt = \frac{1}{4} \int_{\frac12}^{1} t \arctan t \; dt \] But at this point by checking the integral with wolfram alpha the result is already different from the one I get using the original integral for some reason, why is that?

Answer 1

Both integrals are the same, and the value is $$ \int t\arctan t dt = \frac{1}{2}t^2\arctan t-\frac{t}{2}+\frac{1}{2}\arctan t $$ and $$ \frac14 \int_{1/2}^1 t\arctan t dt = \frac{1}{4}(\arctan 1 -\frac14 -\frac{5}{8}\arctan\frac12)\approx 0.061404602 $$