Question Number 32360 by prof Abdo imad last updated on 23/Mar/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{3}\right)}\:. \\ $$
Commented by abdo imad last updated on 25/Mar/18
![I = ∫_0 ^∞ (dx/((2x+1)(2x+3))) =(1/2) ∫_0 ^∞ ( (1/(2x+1)) −(1/(2x+3)))dx (1/4) ∫_0 ^∞ ( (2/(2x+1)) −(2/(2x+3)))dx=(1/4)[ln∣((2x+1)/(2x+3))∣]_0 ^∞ I =−(1/4)ln((1/3)) =(1/4)ln(3) .](https://www.tinkutara.com/question/Q32460.png)
$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{3}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\left(\:\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{3}}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\left(\:\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}\:−\frac{\mathrm{2}}{\mathrm{2}{x}+\mathrm{3}}\right){dx}=\frac{\mathrm{1}}{\mathrm{4}}\left[{ln}\mid\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}+\mathrm{3}}\mid\right]_{\mathrm{0}} ^{\infty} \\ $$$${I}\:=−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{3}\right)\:. \\ $$