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

Ex 7.9

Ex 7.9, 1

Ex 7.9, 2

Ex 7.9, 3

Ex 7.9, 4 Important

Ex 7.9, 5

Ex 7.9, 6

Ex 7.9, 7

Ex 7.9, 8 Important

Ex 7.9, 9

Ex 7.9, 10

Ex 7.9, 11 Important

Ex 7.9, 12

Ex 7.9, 13

Ex 7.9, 14 Important

Ex 7.9, 15

Ex 7.9, 16 Important You are here

Ex 7.9, 17 Important

Ex 7.9, 18

Ex 7.9, 19 Important

Ex 7.9, 20 Important

Ex 7.9, 21 (MCQ) Important

Ex 7.9, 22 (MCQ)

Chapter 7 Class 12 Integrals

Serial order wise

Last updated at Dec. 20, 2019 by Teachoo

Ex 7.9, 16 ā«_1^2ā(5š„^2)/(š„^2 + 4š„ + 3) šš„ Let F(š„)=ā«1āć(5š„^2)/(š„^2 + 4š„ + 3) šš„ć =5ā«1āćš„^2/(š„^2 + 4š„ + 3) šš„ć =5ā«1āć(š„^2 + 4š„ + 3 ā 4š„ ā 3)/(š„^2 + 4š„ + 3) šš„ć =5ā«1āć(š„^2 + 4š„ + 3)/(š„^2 + 4š„ + 3) šš„ćā5ā«1āć( (4š„ + 3))/(š„^2 + 4š„ + 3) šš„ć =ā«1āć(5ā(20š„ + 15 )/(š„^2 + 4š„ + 3)) šš„ć =ā«1āć(5ā(20š„ + 15 )/(š„^2 + 3š„ + š„ + 3)) šš„ć =ā«1āć(5ā(20š„ + 15 )/(š„(š„ + 3) + 1(š„ + 3))) šš„ć =ā«1āć(5ā(20š„ + 15 )/((š„ + 3) (š„ + 1))) šš„ć 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„=ā1 20(ā1)+15=A(ā1 + 1) + B(ā1 + 3) ā20+15=AĆ0+B (2) Putting š„=ā1 20(ā1)+15=A(ā1 + 1) + B(ā1 + 3) ā20+15=AĆ0+B (2) ā5=2B B=(ā5)/( 2) Similarly Putting š„=ā3 20(ā3)+15=A(ā3+1)+B(ā3+3) ā60+15=A(ā2) BĆ0 ā45=ā2A A=45/2 Hence ā«1āā((5š„^2)/(š„^2 + 4š„ + 3) " " =ā«1āć(5ā(20š„ + 15 )/(š„^2 + 4š„ + 3)) šš„ć) =ā«1āć5āA/(š„+3)āć B/(š„+1) šš„ =ā«1āć5 šš„ćāā«1āć(45/2)/(š„ +_3) šš„āā«1āć(((ā5)/( 2)))/(š„+1) šš„ćć =5š„ā45/2 ššš|š„+3|+5/2 ššš|š„+1| Hence F(š„)=5š„ā5/2 [9 ššš|š„+3|āššš|š„+1|] Now, ā«_1^2āć(5š„^2)/(š„^2 + 4š„ + 3) šš„=š¹(2)āš¹(1) ć =[5Ć2ā5/2 (9 ššš|2+3|āššš|2+1|)] ā [5 Ć1ā5/2 (9 ššš|1+3|āššš|1+1|)] =10ā5/2 [9 ššš 5āššš 3]ā5+5/2 [9 ššš 4āššš 2] =10ā5ā5/2 [9ššš 5āššš 3ā9ššš 4+ššš 2)] =10ā5ā5/2 [9ššš 5ā9ššš 4ā(ššš 3āššš 2)] =šāš/š (š š„šØš š/šāš„šØš š/š) (log a ā log b = log š/š)