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Question Number 7872 by tawakalitu last updated on 22/Sep/16

∫((x − 8)/(x^2 + 4x + 16)) dx

$$\int\frac{{x}\:−\:\mathrm{8}}{{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:+\:\mathrm{16}}\:{dx} \\ $$

Commented by prakash jain last updated on 23/Sep/16

((x−8)/(x^2 +4x+16))=((x−8)/(x^2 +4x+4+12)) =((x−8)/((x+2)^2 −(i(√(12)))^2 )) =((x−8)/((x+2+i(√(12)))(x+2−i(√(12))))) convert to partial fraction (x−8)((1/(x+2−i(√(12))))−(1/(x+2+i(√(12)))))(1/(2i(√(12)))) (((x−8)/(x−2−i(√(12))))−((x−8)/(x−2+i(√(12)))))(1/(2i(√(12)))) (((x+2−i(√(12))−(10−i(√(12))))/(x+2−i(√(12))))−((x+2+i(√(12))−(10+i(√(12))))/(x+2+i(√(12)))))(1/(2i(√(12)))) ((((10−i(√(12))))/(x−2−i(√(12))))−(((10+i(√(12))))/(x−2+i(√(12)))))(1/(2i(√(12))))

$$\frac{{x}−\mathrm{8}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{16}}=\frac{{x}−\mathrm{8}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}+\mathrm{12}} \\ $$$$=\frac{{x}−\mathrm{8}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\left({i}\sqrt{\mathrm{12}}\right)^{\mathrm{2}} } \\ $$$$=\frac{{x}−\mathrm{8}}{\left({x}+\mathrm{2}+{i}\sqrt{\mathrm{12}}\right)\left({x}+\mathrm{2}−{i}\sqrt{\mathrm{12}}\right)} \\ $$$${convert}\:{to}\:{partial}\:{fraction}\: \\ $$$$\left({x}−\mathrm{8}\right)\left(\frac{\mathrm{1}}{{x}+\mathrm{2}−{i}\sqrt{\mathrm{12}}}−\frac{\mathrm{1}}{{x}+\mathrm{2}+{i}\sqrt{\mathrm{12}}}\right)\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{12}}} \\ $$$$\left(\frac{{x}−\mathrm{8}}{{x}−\mathrm{2}−{i}\sqrt{\mathrm{12}}}−\frac{{x}−\mathrm{8}}{{x}−\mathrm{2}+{i}\sqrt{\mathrm{12}}}\right)\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{12}}} \\ $$$$\left(\frac{{x}+\mathrm{2}−{i}\sqrt{\mathrm{12}}−\left(\mathrm{10}−{i}\sqrt{\mathrm{12}}\right)}{{x}+\mathrm{2}−{i}\sqrt{\mathrm{12}}}−\frac{{x}+\mathrm{2}+{i}\sqrt{\mathrm{12}}−\left(\mathrm{10}+{i}\sqrt{\mathrm{12}}\right)}{{x}+\mathrm{2}+{i}\sqrt{\mathrm{12}}}\right)\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{12}}} \\ $$$$\left(\frac{\left(\mathrm{10}−{i}\sqrt{\mathrm{12}}\right)}{{x}−\mathrm{2}−{i}\sqrt{\mathrm{12}}}−\frac{\left(\mathrm{10}+{i}\sqrt{\mathrm{12}}\right)}{{x}−\mathrm{2}+{i}\sqrt{\mathrm{12}}}\right)\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{12}}} \\ $$$$ \\ $$

Commented by tawakalitu last updated on 22/Sep/16

Thanks so much

$${Thanks}\:{so}\:{much} \\ $$

Commented by sandy_suhendra last updated on 23/Sep/16

I think (x+2)^2 −(i(√(12)))^2 =(x+2+i(√(12)))(x+2−i(√(12)))

$${I}\:{think}\:\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\left({i}\sqrt{\mathrm{12}}\right)^{\mathrm{2}} =\left({x}+\mathrm{2}+{i}\sqrt{\mathrm{12}}\right)\left({x}+\mathrm{2}−{i}\sqrt{\mathrm{12}}\right) \\ $$

Commented by prakash jain last updated on 23/Sep/16

Thanks

$$\mathrm{Thanks} \\ $$

Answered by prakash jain last updated on 02/Oct/16

see comments

$$\mathrm{see}\:\mathrm{comments} \\ $$

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