Misc 21 - Chapter 7 Class 12 Integrals

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Misc 21 Integrate the function (š‘„^2 + š‘„ + 1)/((š‘„ + 1)^2 (š‘„ + 2) ) āˆ«1ā–’怖(š‘„^2 + š‘„ + 1)/((š‘„ + 1)^2 (š‘„ + 2) ) " " š‘‘š‘„怗 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 āˆ«1ā–’(š‘„^2+ š‘„ +1)/(怖(š‘„ + 1)怗^2 (š‘„ + 2))=āˆ«1ā–’(3 š‘‘š‘„)/(š‘„ + 2)āˆ’āˆ«1ā–’(2 š‘‘š‘„)/(š‘„ + 1)+āˆ«1ā–’(1 š‘‘š‘„)/(š‘„ + 1)^2 = 3log |š‘„+2| "ā€“ 2log " |š‘„+1|āˆ’ 1/(š‘„ + 1)+š¶ = "ā€“ 2log " |š’™+šŸ|āˆ’ šŸ/(š’™ + šŸ)+"3log " |š’™+šŸ|+š‘Ŗ

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo