Question

Please help me asap. Got stuck with this question

Answer 1

Answer:

Step-by-step explanation:

$a.\\\frac{d}{dx}(\frac{-1}{1+x^2} )=\frac{d}{dx} [-1(1+x^2)^{-1}]=-1(-1)(1+x^2)^{-2}(2x)=\frac{2x}{(1+x^2)^2 }$

b.

put 1+x²=u

2x dx=du

when x=0,u=1

when x=2,u=1+2²=5

$\int\limits^5_1 {\frac{1}{u}} \, du= ln |u| ,1 \to 5=ln5-ln1=ln 5-0=ln 5$