Chapter 7 Class 12 Integrals
Concept wise

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Ex 7.8, 16 āˆ«_1^2ā–’(5š‘„^2)/(š‘„^2 + 4š‘„ + 3) š‘‘š‘„ Let F(š‘„)=āˆ«1ā–’怖(5š‘„^2)/(š‘„^2 + 4š‘„ + 3) š‘‘š‘„怗 =5āˆ«1ā–’怖š’™^šŸ/(š‘„^2 + 4š‘„ + 3) š‘‘š‘„怗 =5āˆ«1ā–’怖(š’™^šŸ + šŸ’š’™ + šŸ‘ āˆ’ šŸ’š’™ āˆ’ šŸ‘)/(š’™^šŸ + šŸ’š’™ + šŸ‘) š‘‘š‘„怗 =5āˆ«1ā–’怖(š‘„^2 + 4š‘„ + 3)/(š‘„^2 + 4š‘„ + 3) š‘‘š‘„怗āˆ’5āˆ«1ā–’怖( (4š‘„ + 3))/(š‘„^2 + 4š‘„ + 3) š‘‘š‘„怗 =āˆ«1ā–’怖(šŸ“āˆ’(šŸšŸŽš’™ + šŸšŸ“ )/(š’™^šŸ + šŸ’š’™ + šŸ‘)) š’…š’™ć€— =āˆ«1ā–’怖(5āˆ’(20š‘„ + 15 )/(š‘„^2 + 3š‘„ + š‘„ + 3)) š‘‘š‘„怗 =āˆ«1ā–’怖(5āˆ’(20š‘„ + 15 )/(š‘„(š‘„ + 3) + 1(š‘„ + 3))) š‘‘š‘„怗 =āˆ«1ā–’怖(šŸ“āˆ’(šŸšŸŽš’™ + šŸšŸ“ )/((š’™ + šŸ‘) (š’™ + šŸ))) š’…š’™ć€— Now, Let (šŸšŸŽš’™ + šŸšŸ“)/(š’™ + šŸ‘)(š’™ + šŸ) =š€/(š’™ + šŸ‘)+š/(š’™ + šŸ) (20š‘„ + 15)/(š‘„ + 3)(š‘„ + 1) =(A(š‘„ + 1) + B(š‘„ + 3))/(š‘„ + 3)(š‘„ + 1) Canceling denominator 20š‘„+15=A(š‘„ + 1) + B(š‘„ + 3) Putting š’™=āˆ’šŸ 20(āˆ’1)+15=A(āˆ’1 + 1) + B(āˆ’1 + 3) āˆ’20+15=AƗ0+B (2) āˆ’5=2B š=(āˆ’šŸ“)/( šŸ) Putting š’™=āˆ’šŸ‘ 20(āˆ’3)+15=A(āˆ’3+1)+B(āˆ’3+3) āˆ’60+15=A(āˆ’2) BƗ0 āˆ’45=āˆ’2A š‘Ø=šŸ’šŸ“/šŸ Hence āˆ«1ā–’ā–ˆ((šŸ“š’™^šŸ)/(š’™^šŸ + šŸ’š’™ + šŸ‘) " " =āˆ«1ā–’怖(5āˆ’(20š‘„ + 15 )/(š‘„^2 + 4š‘„ + 3)) š‘‘š‘„怗) =āˆ«1ā–’怖5āˆ’A/(š‘„ + 3)āˆ’怗 B/(š‘„ + 1) š‘‘š‘„ =āˆ«1ā–’怖šŸ“ š’…š’™ć€—āˆ’āˆ«1ā–’怖(šŸ’šŸ“/šŸ)/(š’™ + šŸ‘) š’…š’™āˆ’āˆ«1ā–’怖(((āˆ’šŸ“)/( šŸ)))/(š’™ + šŸ) š’…š’™ć€—怗 =5š‘„āˆ’45/2 š‘™š‘œš‘”|š‘„+3|+5/2 š‘™š‘œš‘”|š‘„+1| Hence F(š’™)=šŸ“š’™āˆ’šŸ“/šŸ [šŸ— š’š’š’ˆ|š’™+šŸ‘|āˆ’š’š’š’ˆ|š’™+šŸ|] Now, āˆ«_1^2ā–’怖(šŸ“š’™^šŸ)/(š’™^šŸ + šŸ’š’™ + šŸ‘) š’…š’™=š¹(2)āˆ’š¹(1) 怗 =[5 Ɨ 2āˆ’5/2 (9 š‘™š‘œš‘”|2+3|āˆ’š‘™š‘œš‘”|2+1|)] āˆ’ [5 Ɨ 1āˆ’5/2 (9 š‘™š‘œš‘”|1+3|āˆ’š‘™š‘œš‘”|1+1|)] =10āˆ’5/2 [9 š‘™š‘œš‘” 5āˆ’š‘™š‘œš‘” 3]āˆ’5+5/2 [9 š‘™š‘œš‘” 4āˆ’š‘™š‘œš‘” 2] =10āˆ’5āˆ’5/2 [9š‘™š‘œš‘” 5āˆ’š‘™š‘œš‘” 3āˆ’9š‘™š‘œš‘” 4+š‘™š‘œš‘” 2)] =10āˆ’5āˆ’5/2 [9š‘™š‘œš‘” 5āˆ’9š‘™š‘œš‘” 4āˆ’(š‘™š‘œš‘” 3āˆ’š‘™š‘œš‘” 2)] =šŸ“āˆ’šŸ“/šŸ (šŸ— š„šØš  šŸ“/šŸ’āˆ’š„šØš  šŸ‘/šŸ)

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.