Given the improper integral: $$\int_1^\infty 45\frac{x+1}{x^2+2x} \, dx$$ I was able to set up the limits as shown below, but I am not sure how to continue integrating. $$\lim_{t\to\infty}\int_1^t 45\frac{x+1}{x^2+2x} \, dx = \lim_{t\to\infty} 45 \int_1^t \frac{x+1}{x^2+2x} \, dx$$
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$$\begin{align} u & = x^2+2x \ du & = (2x+2) , dx \ du/2 & = (x+1) , dx \end{align}$$ $$ \int_1^t \frac{x+1}{x^2+2x},dx = \int_3^{t^2+2t} \frac{du/2} u $$ – Michael Hardy May 04 '16 at 17:36
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Easiest way to integrate is using partial fractions $\frac{x+1}{x^2+2x}=\frac{1}{2x}+\frac{1}{2(2+x)}$ – almagest May 04 '16 at 17:40
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Doesn't it diverge? – jim May 04 '16 at 17:45
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Thanks, and yes it does diverge. – Derek May 04 '16 at 17:56
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In which case there is no answer – jim May 04 '16 at 17:59
