Question Number 14274 by christine last updated on 30/May/17
Answered by Tinkutara last updated on 30/May/17
![∫_1 ^∞ ((xdx)/(x + 1)) = ∫_1 ^∞ (1 − (1/(x + 1)))dx = ∫_1 ^∞ dx − ∫_1 ^∞ (dx/(x + 1)) = [x]_1 ^∞ − ln ∣x + 1∣_1 ^∞ = ∞ − [ln ∞ − ln 2] = ∞](https://www.tinkutara.com/question/Q14276.png)
$$\int_{\mathrm{1}} ^{\infty} \frac{{xdx}}{{x}\:+\:\mathrm{1}}\:=\:\int_{\mathrm{1}} ^{\infty} \left(\mathrm{1}\:−\:\frac{\mathrm{1}}{{x}\:+\:\mathrm{1}}\right){dx} \\ $$$$=\:\int_{\mathrm{1}} ^{\infty} {dx}\:−\:\int_{\mathrm{1}} ^{\infty} \frac{{dx}}{{x}\:+\:\mathrm{1}} \\ $$$$=\:\left[{x}\right]_{\mathrm{1}} ^{\infty} \:−\:\mathrm{ln}\:\mid{x}\:+\:\mathrm{1}\mid_{\mathrm{1}} ^{\infty} \\ $$$$=\:\infty\:−\:\left[\mathrm{ln}\:\infty\:−\:\mathrm{ln}\:\mathrm{2}\right]\:=\:\infty \\ $$