Answer:
Step-by-step explanation:
[tex]a.\\\frac{d}{dx}(\frac{-1}{1+x^2} )=\frac{d}{dx} [-1(1+x^2)^{-1}]=-1(-1)(1+x^2)^{-2}(2x)=\frac{2x}{(1+x^2)^2 }\\[/tex]
b.
put 1+x²=u
2x dx=du
when x=0,u=1
when x=2,u=1+2²=5
[tex]\int\limits^5_1 {\frac{1}{u}} \, du= ln |u| ,1 \to 5=ln5-ln1=ln 5-0=ln 5[/tex]
