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x-6-64-4-2x-1-x-2-x-2-4-4x-1-x-2-4x-2-2x-1-1-2x-dx-

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Question Number 171926 by Mikenice last updated on 21/Jun/22
∫ (((x^(−6) −64)/(4+2x^(−1) +x^(−2) )).(x^2 /(4−4x^(−1) +x^(−2) )) − ((4x^2 (2x+1))/(1−2x)))dx
$$\int\:\left(\frac{{x}^{−\mathrm{6}} −\mathrm{64}}{\mathrm{4}+\mathrm{2}{x}^{−\mathrm{1}} +{x}^{−\mathrm{2}} }.\frac{{x}^{\mathrm{2}} }{\mathrm{4}−\mathrm{4}{x}^{−\mathrm{1}} +{x}^{−\mathrm{2}} }\:−\:\frac{\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{1}−\mathrm{2}{x}}\right){dx} \\ $$
Commented by infinityaction last updated on 22/Jun/22
x^2 +x+c
$${x}^{\mathrm{2}} +{x}+{c} \\ $$
Commented by infinityaction last updated on 22/Jun/22
p = ((x^(−6) −64)/(4+2x^(−1) +x^(−2) )).(x^2 /(4−4x^(−1) +x^(−2) )) − ((4x^2 (2x+1))/(1−2x)) p = (((1/x^6 )−64)/(4+(2/x)+(1/x^2 ))) ∙(x^2 /(4−(4/x)+(1/x^2 )))−((4x^2 )/(1−2x)) p = (((1−64x^4 )x^2 )/(x^6 (4x^2 +2x+1)))∙ (x^4 /((4x^2 −4x+1)))−((4x^2 (1+2x))/(1−2x)) 1−64x^6 = 1^3 −(4x^2 )^3 1−64x^6 = (1−4x^2 )(1+16x^4 +4x^2 ) p = (((1−4x^2 )(1+16x^4 +4x^2 ))/((4x^2 +2x+1))∙(1/(4x^2 −4x+1))−((4x^2 (2x+1))/(1−2x)) p = ((−(2x−1)(1+2x)(1+16x^4 4x^2 ))/(4x^2 +2x+1)(2x−1)^2 ))∙+((4x^2 (1+2x))/(2x−1)) p = ((2x+1)/(2x−1)){−((1+16x^4 +4x2)/(1+2x+4x^2 )) +4x^2 } p = ((2x+1)/(2x−1)){ ((−1−16x^4 −4x^2 +8x^3 +4x^2 +16x^4 )/(1+2x+4x^2 ))} p = ((2x+1)/(2x−1)){(((2x)^3 −1)/(1+2x+4x^2 ))} p = ((2x+1)/(2x−1)){(((2x−1) (1+2x+4x^2 ))/(1+2x+4x^2 ))} p = 2x+1 ∫pdx = ∫(2x+1)dx ∫pdx = x^2 +x+C
$$\boldsymbol{{p}}\:=\:\:\frac{\boldsymbol{{x}}^{−\mathrm{6}} −\mathrm{64}}{\mathrm{4}+\mathrm{2}\boldsymbol{{x}}^{−\mathrm{1}} +\boldsymbol{{x}}^{−\mathrm{2}} }.\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{4}−\mathrm{4}\boldsymbol{{x}}^{−\mathrm{1}} +\boldsymbol{{x}}^{−\mathrm{2}} }\:−\:\frac{\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \left(\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)}{\mathrm{1}−\mathrm{2}\boldsymbol{{x}}} \\ $$$$\:\:\boldsymbol{{p}}\:=\:\frac{\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{6}} }−\mathrm{64}}{\mathrm{4}+\frac{\mathrm{2}}{\boldsymbol{{x}}}+\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} }}\:\centerdot\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{4}−\frac{\mathrm{4}}{\boldsymbol{{x}}}+\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} }}−\frac{\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}\boldsymbol{{x}}} \\ $$$$\:\:\boldsymbol{{p}}\:\:=\:\:\:\frac{\left(\mathrm{1}−\mathrm{64}\boldsymbol{{x}}^{\mathrm{4}} \right)\cancel{\boldsymbol{{x}}^{\mathrm{2}} }}{\cancel{\boldsymbol{{x}}^{\mathrm{6}} }\left(\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)}\centerdot\:\frac{\cancel{\boldsymbol{{x}}^{\mathrm{4}} }}{\left(\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}+\mathrm{1}\right)}−\frac{\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}\boldsymbol{{x}}\right)}{\mathrm{1}−\mathrm{2}\boldsymbol{{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{1}−\mathrm{64}{x}^{\mathrm{6}} \:=\:\mathrm{1}^{\mathrm{3}} −\left(\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\mathrm{1}−\mathrm{64}{x}^{\mathrm{6}} \:=\:\left(\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{16}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} \right) \\ $$$$\:\boldsymbol{{p}}\:\:=\:\:\:\:\frac{\left(\mathrm{1}−\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{16}\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \right)}{\left(\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right.}\centerdot\frac{\mathrm{1}}{\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}+\mathrm{1}}−\frac{\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \left(\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)}{\mathrm{1}−\mathrm{2}\boldsymbol{{x}}} \\ $$$$\boldsymbol{{p}}\:\:=\:\:\:\frac{−\left(\mathrm{2}\boldsymbol{{x}}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{2}\boldsymbol{{x}}\right)\left(\mathrm{1}+\mathrm{16}\boldsymbol{{x}}^{\mathrm{4}} \mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \right)}{\left.\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)\left(\mathrm{2}\boldsymbol{{x}}−\mathrm{1}\right)^{\mathrm{2}} }\centerdot+\frac{\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}\boldsymbol{{x}}\right)}{\mathrm{2}\boldsymbol{{x}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:{p}\:\:\:\:=\:\:\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\left\{−\frac{\mathrm{1}+\mathrm{16}{x}^{\mathrm{4}} +\mathrm{4}{x}\mathrm{2}}{\mathrm{1}+\mathrm{2}{x}+\mathrm{4}{x}^{\mathrm{2}} }\:\:+\mathrm{4}{x}^{\mathrm{2}} \right\} \\ $$$$\:{p}\:\:=\:\:\frac{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}}{\mathrm{2}\boldsymbol{{x}}−\mathrm{1}}\left\{\:\frac{−\mathrm{1}−\cancel{\mathrm{16}\boldsymbol{{x}}^{\mathrm{4}} }−\cancel{\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} }+\mathrm{8}\boldsymbol{{x}}^{\mathrm{3}} +\cancel{\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} }+\cancel{\mathrm{16}\boldsymbol{{x}}^{\mathrm{4}} }}{\mathrm{1}+\mathrm{2}\boldsymbol{{x}}+\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} }\right\} \\ $$$$\:\:\:\:\:{p}\:=\:\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\left\{\frac{\left(\mathrm{2}{x}\right)^{\mathrm{3}} −\mathrm{1}}{\mathrm{1}+\mathrm{2}{x}+\mathrm{4}{x}^{\mathrm{2}} }\right\} \\ $$$$\:\:\:\:{p}\:\:=\:\:\frac{\mathrm{2}{x}+\mathrm{1}}{\cancel{\mathrm{2}{x}−\mathrm{1}}}\left\{\frac{\cancel{\left(\mathrm{2}{x}−\mathrm{1}\right)}\:\cancel{\left(\mathrm{1}+\mathrm{2}{x}+\mathrm{4}{x}^{\mathrm{2}} \right)}}{\cancel{\mathrm{1}+\mathrm{2}{x}+\mathrm{4}{x}^{\mathrm{2}} }}\right\} \\ $$$$\:\:\:\:\:\:\:\:{p}\:\:=\:\mathrm{2}{x}+\mathrm{1} \\ $$$$\:\:\:\:\:\int{pdx}\:\:=\:\:\int\left(\mathrm{2}{x}+\mathrm{1}\right){dx} \\ $$$$\:\:\:\:\:\:\int{pdx}\:\:=\:\:\:{x}^{\mathrm{2}} +{x}+{C} \\ $$
Commented by Mikenice last updated on 23/Jun/22
thanks sir
$${thanks}\:{sir} \\ $$

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