Question Number 104505 by mathmax by abdo last updated on 22/Jul/20
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$
Answered by OlafThorendsen last updated on 22/Jul/20
![R(x) = (1/((x+1)^2 (x+2)^2 (x+3)^2 )) R(x) = (1/(4(x+1)^2 ))+(1/((x+2)^2 ))+(1/(4(x+3)^2 )) −(3/(4(x+1)))+(3/(4(x+3))) ∫_0 ^∞ R(x)dx = [−(1/(4(x+1)))−(1/(x+2))−(1/(4(x+3)))−(3/4)ln∣((x+1)/(x+3))∣]_0 ^∞ = (1/4)+(1/2)+(1/(12))+(3/4)ln(1/3) = (5/6)−(3/4)ln3](https://www.tinkutara.com/question/Q104508.png)
$$\mathrm{R}\left({x}\right)\:=\:\:\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{R}\left({x}\right)\:=\:\:\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{3}}{\mathrm{4}\left(\mathrm{x}+\mathrm{1}\right)}+\frac{\mathrm{3}}{\mathrm{4}\left(\mathrm{x}+\mathrm{3}\right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{R}\left(\mathrm{x}\right)\mathrm{dx}\:= \\ $$$$\left[−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{x}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{x}+\mathrm{3}\right)}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln}\mid\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}+\mathrm{3}}\mid\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln}\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=\:\frac{\mathrm{5}}{\mathrm{6}}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln3} \\ $$