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iHav to Drive
Eastern Imports
How do you integrate (-infinity to 5) 1/(x^2+1)dx?
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<blockquote data-quote="CherryB" data-source="post: 1657318" data-attributes="member: 477112"><p>When I integrate </p><p>int (-infinity to 5) 1/(x^2+1)dx by using the limit as t --> -infinity of int (t to 5) 1/(x^2+1)dx, I get negative infinity, which means it is DIVERGENT. (antiderivative of 1/(x^2+1) is arctan x, right? and since arctan t as t--> -infinity is -infinity, I get divergent)</p><p></p><p>But the answer gives a numerical value for the answer, which means it is convergent... why?</p></blockquote><p></p>
[QUOTE="CherryB, post: 1657318, member: 477112"] When I integrate int (-infinity to 5) 1/(x^2+1)dx by using the limit as t --> -infinity of int (t to 5) 1/(x^2+1)dx, I get negative infinity, which means it is DIVERGENT. (antiderivative of 1/(x^2+1) is arctan x, right? and since arctan t as t--> -infinity is -infinity, I get divergent) But the answer gives a numerical value for the answer, which means it is convergent... why? [/QUOTE]
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