Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 14667 by Umar math last updated on 03/Jun/17

Commented by prakash jain last updated on 03/Jun/17

(1/((x^3 −1)^3 ))=(1/((x−1)^3 (x^2 +x+1)^3 ))  =(A/(x−1))+(B/((x−1)^2 ))+(C/((x−1)^3 ))  =((Dx+E)/(x^2 +x+1))+((Fx+G)/((x^2 +x+1)^2 ))+((Hx+I)/((x^2 +x+1)^3 ))  computing  (5/(27(x−1)))−(1/(9(x−1)^2 ))+(1/(27(x−1)^3 ))  +((−2x−1)/(9(x^2 +x+1)^3 ))+((−7x−8)/(27(x^2 +x+1)^2 ))+((−5x−7)/(27(x^2 +x+1)))  Each of the term in the partial  fraction can be integrated easily.  However this is a lengthy procedure.

$$\frac{\mathrm{1}}{\left({x}^{\mathrm{3}} −\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{{A}}{{x}−\mathrm{1}}+\frac{{B}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{C}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{{Dx}+{E}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\frac{{Fx}+{G}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{Hx}+{I}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${computing} \\ $$$$\frac{\mathrm{5}}{\mathrm{27}\left({x}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{9}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{27}\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$+\frac{−\mathrm{2}{x}−\mathrm{1}}{\mathrm{9}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{3}} }+\frac{−\mathrm{7}{x}−\mathrm{8}}{\mathrm{27}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{−\mathrm{5}{x}−\mathrm{7}}{\mathrm{27}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)} \\ $$$$\mathrm{Each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:\mathrm{the}\:\mathrm{partial} \\ $$$$\mathrm{fraction}\:\mathrm{can}\:\mathrm{be}\:\mathrm{integrated}\:\mathrm{easily}. \\ $$$$\mathrm{However}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{lengthy}\:\mathrm{procedure}. \\ $$

Commented by Umar math last updated on 04/Jul/17

thanks sir.

$${thanks}\:{sir}. \\ $$

Answered by Umar math last updated on 04/Jun/17

thank u sir

$${thank}\:{u}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com