Question

решить несобственный интеграл,ПОЖАЛУЙСТА
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Answer 1

$\displaystyle\\\int\limits^\infty_1 {\frac{dx}{(x^2+1)^2} } \, dx\\\\\\\int\limits^\infty_a {f(x)} \, dx= \lim_{b \to +\infty} F(x)\mid^b_a\\\\\\ \int \frac{dx}{(x^2+1)^2}=\frac{x}{2(x^2+1)}+\frac{1}{2}\int\frac{1}{x^2+1}dx=\frac{x}{2(x^2+1)}+\frac{1}{2}\arctan(x)\\\\\\ \lim_{b \to +\infty} (\frac{x}{2(x^2+1)}+\frac{1}{2}\arctan(x))\mid^b_1= \lim_{b \to +\infty} (\frac{b}{2(b^2+1)}+\frac{1}{2}\arctan(b)-\\\\\\-(\frac{1}{2(1^2+1)}+\frac{1}{2}\arctan(1)))=$

$\displaystyle\\=\lim_{b \to +\infty} (\frac{b}{2(b^2+1)}+\frac{1}{2}\arctan(b)-(\frac{1}{4}+\frac{\pi}{8}))= \lim_{b \to +\infty}(\frac{\frac{1}{b} }{2+\frac{2}{b^2} }+\frac{1}{2}\arctan(b)-\\\\\\ -(\frac{1}{4}+\frac{\pi}{8}))=0+\frac{1}{2}*\frac{\pi}{2}-(\frac{1}{4}+\frac{\pi}{8})=\frac{\pi}{8}-\frac{1}{4}$