Check sibling questions

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

Solve all your doubts with Teachoo Black (new monthly pack available now!)


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‘)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.