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Question Number 208062 by mathzup last updated on 03/Jun/24

$$find{\int }_4^{\mathrm{\infty }}\frac{dx}{{(x+1)}^3{(x-3)}^5}$$

Answered by Frix last updated on 03/Jun/24

$$\mathrm{Use}{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}(\mathrm{search}\mathrm{it}\mathrm{on}$$$$\mathrm{the}www)$$$$\int \frac{dx}{{(x+1)}^3{(x-3)}^5}=\frac{15}{4096}\int \frac{dx}{(x+1)(x+3)}+$$$$+\frac{15x^5-135x^4+350x^3+10x^2-877x+125}{4096{(x+1)}^2{(x-3)}^4}=$$$$=\frac{15\ln \mid \frac{x-3}{x+1}\mid }{16384}+\frac{15x^5-135x^4+350x^3+10x^2-877x+125}{4096{(x+1)}^2{(x-3)}^4}+C$$$$\Rightarrow$$$$\mathrm{Answer}\mathrm{is}\frac{23}{102400}+\frac{15\ln 5}{16374}$$

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