Question

Let $F\left(x\right)={\int }_0^x=\frac{2t+1}{t^2-2t+2}dt,x\in [-1,1]$ then

Answer 1

$F^{\prime }\left(x\right)=\frac{2x+1}{x^2-2x+2}$
since $x^2-2x+2=(x-1{)}^2+1>0$

so $F^{\prime }\left(x\right)<0$ for $x<-1/2$

Hence $F$ increases on $(-1/2,1)$ and decreases on $[-1,-1/2]$

The greatest value of f is $F\left(1\right)$ and least value is $F(-1/2)$ But
$F\left(1\right)={\int }_0^1\frac{2t+1}{t^2-2t+2}dt={\int }_0^1\frac{2t-2+3}{t^2-2t+2}dt$
$=\log {|t^2-2t+2|}_0^1+3{\int }_0^1\frac{dt}{(t-1{)}^2+1}$
${=0-\log 2+3{\tan }^{-1}(t-1)|}_0^1$
$=-\log 2+3\frac{\pi }{4}<2$

So $F\left(x\right)<2$ for all $x\in [-1,1]$
$F(-1/2)=\log |t^2-2t+2|{{}_0^{-1/2}+3{\tan }^{-1}(t-1)|}_0^{-1/2}$
$=\log \frac{13}{4}-\log 2+3\left({\tan }^{-1}\right(-3/2)-{\tan }^{-1}(-1\left)\right)$
$=\log 13-\log 8+3{\tan }^{-1}(-1/5)>2-1-3{\tan }^{-1}1=1-\frac{3\pi }{4}$