Question

limita x tinde la infinit din integrala de de la 0 la xdin ln[1+t] dt/x^2
deci e o fractie integrala la numarator si x^2 la numitor

Answer 1

Răspuns:

Explicație pas cu pas:

$\displaystyle\int_0^x\ln(1+t)dt=\int_0^xt'\cdot\ln(1+t)dt=t\cdot\ln(1+t)|_0^x-\int_0^x\dfrac{t}{1+t}dt =\\x\cdot\ln(1+x)-\int_0^x\dfrac{t+1-1}{t+1}dt=x\cdot\ln(x+1)-\int_0^xdt+\int_0^x\dfrac{1}{t+1}dt=\\=x\cdot\ln(x+1)-t|_0^x+\ln(1+t)|_0^x=x\cdot\ln(x+1)-x+\ln(x+1)\\\lim_{x\to\infty}\dfrac{\int_0^x\ln(1+t)dt}{x^2}=\lim_{x\to\infty}\dfrac{x\cdot\ln(x+1)-x+\ln(x+1)}{x^2}=\\=\lim_{x\to\infty}\left(\dfrac{\ln(1+x)}{x}-\dfrac{1}{x}+\dfrac{\ln(x+1)}{x^2}\right)=0-0+0=\boxed{0}$