If $I=\int_{1 / 2}^2 \frac{\ln t}{1+t^n} d t, \quad(t
eq 1)$

Question

If $I=\int_{1 / 2}^2 \frac{\ln t}{1+t^n} d t, \quad(t 
eq 1)$

Professor PC Thomas Classes · Answer

$ \begin{aligned} & I=\int_{1 / 2}^2 \frac{\ln t}{1+t^n} d t \ & 	ext { put } \quad t=\frac{1}{u} \Rightarrow \quad d t=-\frac{1}{u^2} d u \ & I=\int_2^{1 / 2} \frac{\ln (1 / u)}{1+\left(1 / u^night)}\left(-\frac{1}{u^2}ight) d u=-\int_{1 / 2}^2 \frac{u^n \ln u}{(1+u)^n u^2} d u \ & I=-\int_{1 / 2}^2 \frac{t^{n-2} \ln t}{1+t^n} d t \end{aligned} $ 
$ 2 I=\int_{1 / 2}^2 \frac{\left(1-t^{n-2}ight) \ln t}{\left(1+t^night)} d t $ 
for $n=1$, the integrand is $+v e$ in $(1 / 2,1) \cup(1,2)$ and zero only for $t=1 \Rightarrow 1>0$ 
for $n=2$, integrand is zero $\Rightarrow I=0$ 
for $n \geq 3$, integrand is always $-v e$ except for $t=1$ hence $I<0$ 
Note that for $n=0 I=\int_{1 / 2}^2 \ln t d t=(t \ln t-t)_{y / 2}^2$
$\left.=(2 \ln 2-2)-\left(\frac{1}{2} \ln \frac{1}{2}-\frac{1}{2}ight)=\frac{5}{2} \ln 2-\frac{3}{2}>0ight]$