Misc 21 - Integrate x^2 + x + 1 / (x + 1)^2 (x + 2) - Class 12 - Miscellaneous

part 2 - Misc 21 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals

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Misc 21 Integrate the function (š‘„^2 + š‘„ + 1)/((š‘„ + 1)^2 (š‘„ + 2) ) ∫1▒〖(š‘„^2 + š‘„ + 1)/((š‘„ + 1)^2 (š‘„ + 2) ) " " š‘‘š‘„ć€— By partial fraction (š‘„^2 + š‘„ + 1)/((š‘„ + 1)^2 (š‘„ + 2) )=A/(š‘„ + 2)+B/(š‘„ + 1)+C/怖(š‘„ + 1)怗^2 (š‘„^2 + š‘„ + 1)/((š‘„ + 1)^2 (š‘„ + 2) )=(A怖(š‘„ + 1)怗^2 + B(š‘„ + 1)(š‘„ + 2) + C(š‘„ + 2))/(怖(š‘„ + 1)怗^2 (š‘„ + 2) ) Cancelling denominators š‘„^2+š‘„+1=A怖 (š‘„+1)怗^2+B(š‘„+2)(š‘„+1)+C(š‘„+2) Hence, (š‘„^2 + š‘„ + 1)/((š‘„ + 1)^2 (š‘„ + 2))=3/(š‘„ +2)āˆ’2/(š‘„ +1)+1/怖(š‘„ + 1)怗^2 ∫1ā–’(š‘„^2+ š‘„ +1)/(怖(š‘„ + 1)怗^2 (š‘„ + 2))=∫1ā–’(3 š‘‘š‘„)/(š‘„ + 2)āˆ’āˆ«1ā–’(2 š‘‘š‘„)/(š‘„ + 1)+∫1ā–’(1 š‘‘š‘„)/(š‘„ + 1)^2 = 3log |š‘„+2| "– 2log " |š‘„+1|āˆ’ 1/(š‘„ + 1)+š¶ = "– 2log " |š’™+šŸ|āˆ’ šŸ/(š’™ + šŸ)+"3log " |š’™+šŸ|+š‘Ŗ

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