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Question Number 104505 by mathmax by abdo last updated on 22/Jul/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{x}+1)}^2{(\mathrm{x}+2)}^2{(\mathrm{x}+3)}^2}$$

Answered by OlafThorendsen last updated on 22/Jul/20

$$\mathrm{R}\left(x\right)=\frac{1}{{(\mathrm{x}+1)}^2{(\mathrm{x}+2)}^2{(\mathrm{x}+3)}^2}$$$$\mathrm{R}\left(x\right)=\frac{1}{4{(\mathrm{x}+1)}^2}+\frac{1}{{(\mathrm{x}+2)}^2}+\frac{1}{4{(\mathrm{x}+3)}^2}$$$$-\frac{3}{4(\mathrm{x}+1)}+\frac{3}{4(\mathrm{x}+3)}$$$${\int }_0^{\mathrm{\infty }}\mathrm{R}\left(\mathrm{x}\right)\mathrm{dx}=$$$${[-\frac{1}{4(\mathrm{x}+1)}-\frac{1}{\mathrm{x}+2}-\frac{1}{4(\mathrm{x}+3)}-\frac{3}{4}\ln \mid \frac{\mathrm{x}+1}{\mathrm{x}+3}\mid ]}_0^{\mathrm{\infty }}$$$$=\frac{1}{4}+\frac{1}{2}+\frac{1}{12}+\frac{3}{4}\ln \frac{1}{3}$$$$=\frac{5}{6}-\frac{3}{4}\ln 3$$

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