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Question Number 86638 by M±th+et£s last updated on 29/Mar/20

$$I=\int \frac{x^3-2x^2+7x-1}{{(x-3)}^3{(x-2)}^2}dx$$

Answered by MJS last updated on 29/Mar/20

$$\frac{x^3-2x^2+7x-1}{{(x-3)}^3{(x-2)}^2}=$$$$=\frac{29}{{(x-3)}^3}-\frac{36}{{(x-3)}^2}+\frac{50}{x-3}-\frac{13}{{(x-2)}^2}-\frac{50}{x-2}$$$$\Rightarrow$$$$I=\frac{98x^2+545x+724}{2{(x-3)}^2(x-2)}+50\ln \mid \frac{x-3}{x-2}\mid +C$$$$\mathrm{or}$$$$t=\frac{x-3}{x-2}\to dx={(x-2)}^{2}dt$$$$\Rightarrow$$$$I=-13\int dt+50\int \frac{dt}{t}-65\int \frac{dt}{t^2}+29\int \frac{dt}{t^3}$$$$\mathrm{and}\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}$$

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