Chapter 7 Class 12 Integrals
Concept wise

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.