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Question Number 142149 by cesarL last updated on 27/May/21

$$6.\int \frac{dx}{x^2-3x+2}$$$$7.\int \frac{4dx}{x^2+2x+4}$$$$8.\int \frac{3-2xdx}{x^2-64}$$$$9.\int \frac{3x-1}{x^3+5x^2+6x}dx$$$$10.\int \frac{4-3x}{x^3-2x}dx$$$$11.\int \frac{dx}{x^3-2x+x}$$$$$$

Answered by mathmax by abdo last updated on 27/May/21

$$1)\mathrm{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2-3\mathrm{x}+2}$$$${\mathrm{x}}^2-3\mathrm{x}+2=0\to \mathrm{\Delta }=9-8=1\Rightarrow {\mathrm{x}}_1=\frac{3+1}{2}=2\mathrm{and}{\mathrm{x}}_2=\frac{3-1}{2}=1\Rightarrow$$$$\mathrm{I}=\int \frac{\mathrm{dx}}{(\mathrm{x}-2)(\mathrm{x}-1)}=\int (\frac{1}{\mathrm{x}-2}-\frac{1}{\mathrm{x}-1})\mathrm{dx}=\log \mid \frac{\mathrm{x}-2}{\mathrm{x}-1}\mid +\mathrm{C}$$

Answered by mathmax by abdo last updated on 27/May/21

$$2)\mathrm{J}=\int \frac{4\mathrm{dx}}{{\mathrm{x}}^2+2\mathrm{x}+4}\Rightarrow \mathrm{J}=\int \frac{4\mathrm{dx}}{{(\mathrm{x}+1)}^2+3}$$$${=}_{\mathrm{x}+1=\sqrt{3}\mathrm{y}}\int \frac{4\sqrt{3}\mathrm{dy}}{3({\mathrm{y}}^2+1)}=\frac{4}{\sqrt{3}}\mathrm{arctany}+\mathrm{C}$$$$=\frac{4}{\sqrt{3}}\arctan \left(\frac{\mathrm{x}+1}{\sqrt{3}}\right)+\mathrm{C}$$

Answered by mathmax by abdo last updated on 27/May/21

$$3)\mathrm{I}=\int \frac{-2\mathrm{x}+3}{{\mathrm{x}}^2-64}\mathrm{dx}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{-2\mathrm{x}+3}{{\mathrm{x}}^2-64}=\frac{-2\mathrm{x}+3}{(\mathrm{x}-8)(\mathrm{x}+8)}=\frac{\mathrm{a}}{\mathrm{x}-8}+\frac{\mathrm{b}}{\mathrm{x}+8}$$$$\mathrm{a}=\frac{-16+3}{16}=-\frac{13}{16}$$$$\mathrm{b}=\frac{16+3}{-16}=-\frac{19}{16}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=-\frac{13}{16(\mathrm{x}-8)}-\frac{19}{16(\mathrm{x}+8)}\Rightarrow$$$$\mathrm{I}=-\frac{13}{16}\log \mid \mathrm{x}-8\mid -\frac{19}{16}\log \mid \mathrm{x}+8\mid +\mathrm{C}$$

Answered by mathmax by abdo last updated on 27/May/21

$$5)\mathrm{J}=\int \frac{3\mathrm{x}-1}{{\mathrm{x}}^3+{5\mathrm{x}}^2+6\mathrm{x}}\mathrm{dx}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{3\mathrm{x}-1}{\mathrm{x}({\mathrm{x}}^2+5\mathrm{x}+6)}$$$${\mathrm{x}}^2+5\mathrm{x}+6=0\to \mathrm{\Delta }=25-24=1\Rightarrow {\mathrm{x}}_1=\frac{-5+1}{2}=-2$$$${\mathrm{x}}_2=\frac{-5-1}{2}=-3\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{3\mathrm{x}-1}{\mathrm{x}(\mathrm{x}+2)(\mathrm{x}+3)}=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+2}+\frac{\mathrm{c}}{\mathrm{x}+3}$$$$\mathrm{a}=-\frac{1}{6},\mathrm{b}=\frac{-7}{(-2)\left(1\right)}=\frac{7}{2},\mathrm{c}=\frac{-10}{(-3)(-1)}=-\frac{10}{3}\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=-\frac{1}{6\mathrm{x}}+\frac{7}{2(\mathrm{x}+2)}-\frac{10}{3(\mathrm{x}+3)}\Rightarrow \mathrm{J}=\int \mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}$$$$=-\frac{1}{6}\log \mid \mathrm{x}\mid +\frac{7}{2}\log \mid \mathrm{x}+2\mid -\frac{10}{3}\log \mid \mathrm{x}+3\mid +\mathrm{C}$$

Answered by Panacea last updated on 27/May/21

$$Q11.$$$$\int \frac{dx}{x^3-2x+x}=\int \frac{dx}{x(x+1)(x-1)}$$$$\int (\frac{-1}{x}+\frac{1/2}{x+1}+\frac{1/2}{x-1})dx$$$$ln\frac{1}{x}+\frac{1}{2}ln(x^2-1)$$$$ln\frac{\left(\sqrt{x^2-1}\right)}{x}$$

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