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Question Number 62811 by mathmax by abdo last updated on 25/Jun/19

$$1)find\int \frac{2x^2-1}{(x+1)(x-3)(x^2-x+2)}dx$$$$2)calculate{\int }_5^{+\mathrm{\infty }}\frac{2x^2-1}{(x+1)(x-3)(x^2-x+2)}dx$$

Commented by Prithwish sen last updated on 25/Jun/19

$$1)\frac{{2\mathrm{x}}^2-1}{(\mathrm{x}+1)(\mathrm{x}-3)({\mathrm{x}}^2-\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{\mathrm{x}-3}+\frac{\mathrm{Cx}+\mathrm{D}}{{\mathrm{x}}^2-\mathrm{x}+2}$$$$\mathrm{By}\mathrm{calculating}\mathrm{we}\mathrm{get}$$$$\int [\frac{-1}{16(\mathrm{x}+1)}+\frac{17}{32(\mathrm{x}-3)}-\frac{15(2\mathrm{x}-1)}{32({\mathrm{x}}^2-\mathrm{x}+2)}-\frac{1}{16({\mathrm{x}}^2-\mathrm{x}+2)}]\mathrm{dx}$$$$=-\frac{1}{16}\ln (\mathrm{x}+1)+\frac{17}{32}\ln (\mathrm{x}-3)-\frac{15}{32}\ln ({\mathrm{x}}^2-\mathrm{x}+2)-\frac{1}{8\sqrt{7}}{\tan }^{-1}\frac{2\mathrm{x}-1}{\sqrt{7}}+\mathrm{C}$$$$\mathrm{waiting}\mathrm{for}\mathrm{feedbacks}$$

Commented by mathmax by abdo last updated on 26/Jun/19

$$1)letdecomposeF\left(x\right)=\frac{2x^2-1}{(x+1)(x-3)(x^2-x+2)}$$$$F\left(x\right)=\frac{a}{x+1}+\frac{b}{x-3}+\frac{cx+d}{x^2-x+2}$$$$a={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{1}{(-4)\left(4\right)}=-\frac{1}{16}$$$$b={lim}_{x\to 3}(x-3)F\left(x\right)=\frac{17}{4\left(8\right)}=\frac{17}{32}\Rightarrow F\left(x\right)=\frac{-1}{16(x+1)}+\frac{17}{32(x-3)}+\frac{cx+d}{x^2-x+2}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=-\frac{1}{16}+\frac{17}{32}+c=\frac{15}{32}+c\Rightarrow c=-\frac{15}{32}\Rightarrow$$$$F\left(x\right)=\frac{-1}{16(x+1)}+\frac{17}{32(x-3)}+\frac{-\frac{15}{32}x+d}{x^2-x+2}$$$$F\left(0\right)=\frac{-1}{-6}=\frac{1}{6}=-\frac{1}{16}-\frac{17}{3.32}+\frac{d}{2}\Rightarrow \frac{1}{3}=-\frac{1}{8}-\frac{17}{48}+d\Rightarrow d=\frac{1}{3}+\frac{1}{8}+\frac{17}{48}$$$$=\frac{11}{24}+\frac{17}{48}=\frac{22+17}{48}=\frac{39}{48}\Rightarrow$$$$\int \frac{2x^2-1}{(x+1)(x-3)(x^2-x+2)}dx$$$$=-\frac{1}{16}\int \frac{dx}{x+1}+\frac{17}{32}\int \frac{dx}{x-3}-\frac{15}{32}\int \frac{x}{x^2-x+2}dx+\frac{39}{48}\int \frac{dx}{x^2-x+2}+c$$$$=-\frac{1}{16}ln\mid x+1\mid +\frac{17}{32}ln\mid x-3\mid -\frac{15}{64}\int \frac{2x-1+1}{x^2-x+2}dx+\frac{39}{48}\int \frac{dx}{x^2-x+2}+c$$$$=-\frac{1}{16}ln\mid x+1\mid +\frac{17}{32}ln\mid x-3\mid -\frac{15}{64}ln(x^2-x+2)+(\frac{39}{48}-\frac{15}{64})\int \frac{dx}{x^2-x+2}dx+cbut$$$$\int \frac{dx}{x^2-x+2}=\int \frac{dx}{x^2-x+\frac{1}{4}+2-\frac{1}{4}}=\int \frac{dx}{{(x-\frac{1}{2})}^2+\frac{7}{4}}{=}_{x-\frac{1}{2}=\frac{\sqrt{7}}{2}u}\frac{4}{7}\int \frac{1}{1+u^2}\frac{\sqrt{7}}{2}du$$$$=\frac{2}{\sqrt{7}}arctan\left(\frac{2x-1}{\sqrt{7}}\right)\Rightarrow$$$$\int F\left(x\right)dx=-\frac{1}{16}ln\mid x+1\mid +\frac{17}{32}ln\mid x-3\mid -\frac{15}{64}ln(x^2-x+2)+(\frac{39}{48}-\frac{15}{64})\frac{2}{\sqrt{7}}arctan\left(\frac{2x-1}{\sqrt{7}}\right)+c$$$$$$

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