Question

Am nevoie doar de punctul c, mulțumesc ​

Answer 1

Explicație pas cu pas:

$\int_{0}^{1} \dfrac{ {x}^{2} + 2x + 10}{x + 1} dx = \int_{0}^{1} \dfrac{ {(x + 1)}^{2} + 9}{x + 1} dx = \int_{0}^{1}\Big( x + 1 + \dfrac{9}{x + 1}\Big) dx = \\ = \int_{0}^{1} x \ dx + \int_{0}^{1} 1 \ dx + \int_{0}^{1} \dfrac{9}{x + 1} dx = \dfrac{ {x}^{2} }{2} \Big|_{0}^{1} + x \Big|_{0}^{1} + 9 ln |x + 1| \Big|_{0}^{1}$

$= \dfrac{1}{2} + 1 + 9 ln |2| = \dfrac{3}{2} + 9 ln \ 2$

Answer 2

$f:\mathbb{R}\rightarrow \mathbb{R};~~f(x)=\cfrac{x+1}{x^2+2x+10} \\\\c)~g(x)=\cfrac{1}{f(x)} =\cfrac{1}{\cfrac{x+1}{x^2+2x+10} } =1 \cdot \cfrac{x^2+2x+10}{x+1} = \cfrac{x^2+2x+10}{x+1} \\\\ \displaystyle\int\limits^1_0 \cfrac{x^2+2x+10}{x+1} dx=\int\limits^1_0 \cfrac{(x+1)^2+9}{x+1} dx=\int\limits^1_0 \cfrac{(x+1)^2}{x+1} dx+\int\limits^1_0 \cfrac{9}{x+1} dx=\\\\=\int\limits^1_0 (x+1)dx+\int\limits^1_0 \cfrac{9}{x+1} dx=\int\limits^1_0 xdx+\int\limits^1_0 1dx+9\cdot\int\limits^1_0 \cfrac{1}{x+1}dx=$

$= \cfrac{x^2}{2}\Bigg|_0^1+x\Bigg|_0^1+9ln|x+1 |\Bigg|_0^1=\cfrac{1^2}{2} -\cfrac{0^2}{2} +1-0+9(ln|1+1|-ln|0+1|)=\\\\=\cfrac{1}{2}-0+1+9(ln|2|-ln|1|)=\cfrac{1}{2} +1+9(ln2-0)=\cfrac{1}{2} +1+9ln2=\\\\=\cfrac{1+2+18ln2}{2}=\cfrac{3+18ln2}{2}$