Question

Ma ajuta cineva la a si b ?

Answer 1

$\displaystyle \mathtt{2.~~~f:(0,+\infty)\rightarrow\mathbb{R},~f(x)= \frac{x^2+x+1}{x^2} }\\ \\ \mathtt{a)~\int\limits\left(f(x)- \frac{x+1}{x^2} \right)dx}\\ \\ \mathtt{\int\limits\left( \frac{x^2+x+1}{x^2}- \frac{x+1}{x^2} \right)dx=\int\limits \frac{x^2+x+1-x-1}{x^2}dx =}\\ \\ \mathtt{=\int\limits \frac{x^2}{x^2}dx=\int\limits1dx =x+C}$

$\displaystyle \mathtt{b)~\int\limits^2_1f(x)dx}\\ \\ \mathtt{\int\limits^2_1 \frac{x^2+x+1}{x^2}=\int\limits^2_1 \frac{x^2}{x^2}dx+\int\limits^2_1 \frac{x}{x^2}dx+ \int\limits^2_1 \frac{1}{x^2}dx =}\\ \\ \mathtt{=\int\limits^2_11dx+\int\limits^2_1 \frac{1}{x}dx+\int\limits^2_1 \frac{1}{x^2}dx=x\Bigg|^2_1+ln~x\Bigg|^2_1- \frac{1}{x}\Bigg|^2_1= }\\ \\ \mathtt{=(2-1)+(ln~2-ln~1)-\left( \frac{1}{2}- \frac{1}{1}\right) =1+ln~2+ \frac{1}{2}= \frac{3}{2}+ln~2 }$