Decoding $\int_{\frac 12}^{2} \frac{\ln t}{1+t^n}dt$ 
Let:
$$I=\int_{\frac 12}^{2} \frac{\ln t}{1+t^n}dt$$
Find the sign of $I$ for different values of $n\in\mathbb{N}\cup\{0\}.$
Answer:
$I$ is zero at $n=2$, positive before and negative after.

My Attempt:
I used the substitution $t=\frac 1u$ and obtained:
$$I=-\int_{\frac 12}^{2} \frac{u^{n-2}\ln u}{1+u^n}du$$
From here I was able to observe that $I$ is indeed $0$ at $n=2$ as it would result in $I=-I$.
But I'm unable to comment on any other values of $n$ (except by individually evaluating them).
Thanks in advance.
 A: Let
$$ J_n=\int_{\frac 12}^{2} \frac{\ln t}{1+t^n}dt.$$
You already showed that $J_2=0$.
We now observe that if $n>2$ then
$$t^n<t^2 \text{ for } t\in [1/2, 1)$$
and
$$t^n>t^2 \text{ for } t\in (1, 2].$$
Therefore,
$$J_n=\int_{\frac 12}^{1} \frac{\ln t}{1+t^n}dt
+\int_{\frac 12}^{2} \frac{\ln t}{1+t^n}dt
< \int_{\frac 12}^{1} \frac{\ln t}{1+t^2}dt
+\int_{\frac 12}^{2} \frac{\ln t}{1+t^2}dt
=J_2=0
$$
(Note that $\ln t<0$ for $t\in [1/2,1)$)
Similar for the case $n<2$.
A: Indeed, your substitution is almost there, but we just need to split the integral into two integral:
$$I = \int_{\frac 12}^{2} \frac{\ln t}{1+t^n}dt = \int_{\frac {1}{2}}^{1} \frac{\ln t}{1+t^n}dt + \int_{1}^{2} \frac{\ln t}{1+t^n}dt$$
Now, by substitution $t = \frac{1}{u}$ as you did above for the second integral. Hence:
$$I =  \int_{\frac {1}{2}}^{1} \frac{\ln t}{1+t^n}dt -  \int_{\frac {1}{2}}^{1} \frac{u^{n-2}\ln u}{1+u^n}du = \int_{\frac {1}{2}}^{1}\frac{\left(1-t^{n-2}\right)\ln t}{1+t^n}dt$$
Then for the final expression, we can observe that $n\in \{0,1\}, I > 0$. For $n=2$, we have $I = 0$ as you obtained. Finally, for $n \geq 3$, $I < 0$.
