mat Escreveu:
Calcule integral indo de 2 a infinito de x+3/ (x-1)(x²+1) dx
Desde já, Obrigada!
Desde já, Obrigada!
olá.
\(\\\\\\ \int_{2}^{+\infty}\frac{x+3}{(x-1)*(x^{2}+1)}=\lim_{p\rightarrow +\infty }\int_{2}^{p}\frac{x+3}{(x-1)*(x^{2}+1)}\)
para integrar \(\frac{x+3}{(x-1)*(x^{2}+1)}\) , utilize frações parciais e terá como resposta : \(\\\\\\ \2*ln|x-1|-ln|x^{2}+1|-arc tg(x)\)
então ficamos com:
\(\\\\\\ \int_{2}^{+\infty}\frac{x+3}{(x-1)*(x^{2}+1)}dx=\lim_{p\rightarrow +\infty }(2*ln|p-1|-ln|p^{2}+1|-arc tg(p)+ln(5)+arc tg(2) ) \\\\\\ \int_{2}^{+\infty}\frac{x+3}{(x-1)*(x^{2}+1)}dx=\lim_{p\rightarrow +\infty }(ln|\frac{(x-1)^{2}}{x^{2}+1}|-arc tg(p)+ln(5)+arc tg(2) ) \\\\\\ \int_{2}^{+\infty}\frac{x+3}{(x-1)*(x^{2}+1)}dx=\lim_{p\rightarrow +\infty }(ln|\frac{x^{2}-2x+1}{x^{2}+1}|-arc tg(p)+ln(5)+arc tg(2) ) \\\\\\ \int_{2}^{+\infty}\frac{x+3}{(x-1)*(x^{2}+1)}dx=\lim_{p\rightarrow +\infty }(ln|\frac{x^{2}(1-\frac{2}{x}+\frac{1}{x^{2}})}{x^{2}(1+\frac{1}{x^{2}})}|-arc tg(p)+ln(5)+arc tg(2) ) \\\\\\ \int_{2}^{+\infty}\frac{x+3}{(x-1)*(x^{2}+1)}dx=\lim_{p\rightarrow +\infty }(ln|\frac{1-\frac{2}{x}+\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}}|-arc tg(p)+ln(5)+arc tg(2) ) \\\\\\ \int_{2}^{+\infty}\frac{x+3}{(x-1)*(x^{2}+1)}dx=ln|1|-\frac{\pi}{2}+ln(5)+arc tg(2) \\\\\\ \int_{2}^{+\infty}\frac{x+3}{(x-1)*(x^{2}+1)}dx=-\frac{\pi}{2}+ln(5)+arc tg(2)\)
então a integral converge.