Question

Se consideră funcţia $f:(-1,+\infty) \rightarrow \mathbb{R}, f(x)=x+\frac{2}{x+1}$.

$5 p$ a) Arătați că $\int_{0}^{1}(f(x)-x) d x=2 \ln 2$.

5p b) Calculați $\int_{1}^{e}\left(f(x)-\frac{2}{x+1}\right) \ln x d x$

Answer 1

$f(x)=x+\frac{2}{x+1}$

a)

Folosim tabelul integralelor (vezi atasament)

$\int\limits^1_0 {x+\frac{2}{x+1}-x } \, dx =\int\limits^1_0 {\frac{2}{x+1} } \, dx=2ln(x+1)|_0^1=2ln2-2ln1=2ln2$

b)

$\int\limits^e_1 {(x+\frac{2}{x+2}-\frac{2}{x+1} }) lnx\, dx =\int\limits^e_1 {xlnx} \, dx$

Calculam prin integrarea prin parti

$f=lnx\ \ \ \ \ \ \ \ f'=\frac{1}{x} \\\\g'=x\ \ \ \ \ \ \ \ \ g=\frac{x^2}{2}$

$\int\limits^e_1 {xlnx} \, dx=\frac{x^2}{2}lnx|_1^e-\int\limits^e_1 {\frac{x}{2} } \, dx = \frac{x^2}{2}lnx|_1^e-\frac{x^2}{4} |_1^e=\frac{e^2}{2}-\frac{e^2}{4}+\frac{1}{4} =\frac{e^2+1}{4}$

Un exercitiu similar cu integrale gasesti aici: https://brainly.ro/tema/1026361

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