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Ex 7.9, 16 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Dec. 20, 2019 by

Ex 7.9, 16 - Direct Integrate 5x2 / x2 + 4x + 3 dx from 1 to 2

Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 5

Transcript

Ex 7.9, 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‘„^2/(š‘„^2 + 4š‘„ + 3) š‘‘š‘„怗 =5āˆ«1ā–’怖(š‘„^2 + 4š‘„ + 3 āˆ’ 4š‘„ āˆ’ 3)/(š‘„^2 + 4š‘„ + 3) 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(5āˆ’(20š‘„ + 15 )/(š‘„^2 + 4š‘„ + 3)) š‘‘š‘„怗 =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(5āˆ’(20š‘„ + 15 )/((š‘„ + 3) (š‘„ + 1))) š‘‘š‘„怗 Now, Let (20š‘„ + 15)/(š‘„ + 3)(š‘„ + 1) =A/(š‘„ + 3)+B/(š‘„ + 1) (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’1 20(āˆ’1)+15=A(āˆ’1 + 1) + B(āˆ’1 + 3) āˆ’20+15=AƗ0+B (2) Putting š‘„=āˆ’1 20(āˆ’1)+15=A(āˆ’1 + 1) + B(āˆ’1 + 3) āˆ’20+15=AƗ0+B (2) āˆ’5=2B B=(āˆ’5)/( 2) Similarly Putting š‘„=āˆ’3 20(āˆ’3)+15=A(āˆ’3+1)+B(āˆ’3+3) āˆ’60+15=A(āˆ’2) BƗ0 āˆ’45=āˆ’2A A=45/2 Hence āˆ«1ā–’ā–ˆ((5š‘„^2)/(š‘„^2 + 4š‘„ + 3) " " =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5 š‘‘š‘„怗āˆ’āˆ«1ā–’怖(45/2)/(š‘„ +_3) š‘‘š‘„āˆ’āˆ«1ā–’怖(((āˆ’5)/( 2)))/(š‘„+1) š‘‘š‘„怗怗 =5š‘„āˆ’45/2 š‘™š‘œš‘”|š‘„+3|+5/2 š‘™š‘œš‘”|š‘„+1| Hence F(š‘„)=5š‘„āˆ’5/2 [9 š‘™š‘œš‘”|š‘„+3|āˆ’š‘™š‘œš‘”|š‘„+1|] Now, āˆ«_1^2ā–’怖(5š‘„^2)/(š‘„^2 + 4š‘„ + 3) š‘‘š‘„=š¹(2)āˆ’š¹(1) 怗 =[5Ɨ2āˆ’5/2 (9 š‘™š‘œš‘”|2+3|āˆ’š‘™š‘œš‘”|2+1|)] āˆ’ [5 Ɨ1āˆ’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) (log a āˆ’ log b = log š‘Ž/š‘)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.