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Question Number 167713 by infinityaction last updated on 23/Mar/22

Answered by Nimatullah last updated on 23/Mar/22

∫_0 ^∞ (1/(x+1))dx=ln[x+1]_0 ^∞ =∞

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}+\mathrm{1}}{dx}=\mathrm{ln}\left[{x}+\mathrm{1}\right]_{\mathrm{0}} ^{\infty} =\infty \\ $$

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