Definite Integration - By Partial Fraction

Chapter 7 Class 12 Integrals
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Ex 7.9, 16 β«_1^2β(5π₯^2)/(π₯^2 + 4π₯ + 3) ππ₯ Let F(π₯)=β«1βγ(5π₯^2)/(π₯^2 + 4π₯ + 3) ππ₯γ =5β«1βγπ₯^2/(π₯^2 + 4π₯ + 3) ππ₯γ =5β«1βγ(π₯^2 + 4π₯ + 3 β 4π₯ β 3)/(π₯^2 + 4π₯ + 3) ππ₯γ =5β«1βγ(π₯^2 + 4π₯ + 3)/(π₯^2 + 4π₯ + 3) ππ₯γβ5β«1βγ( (4π₯ + 3))/(π₯^2 + 4π₯ + 3) ππ₯γ =β«1βγ(5β(20π₯ + 15 )/(π₯^2 + 4π₯ + 3)) ππ₯γ =β«1βγ(5β(20π₯ + 15 )/(π₯^2 + 3π₯ + π₯ + 3)) ππ₯γ =β«1βγ(5β(20π₯ + 15 )/(π₯(π₯ + 3) + 1(π₯ + 3))) ππ₯γ =β«1βγ(5β(20π₯ + 15 )/((π₯ + 3) (π₯ + 1))) ππ₯γ Now, Let (20π₯ + 15)/(π₯ + 3)(π₯ + 1) =A/(π₯ + 3)+B/(π₯ + 1) (20π₯ + 15)/(π₯ + 3)(π₯ + 1) =(A(π₯ + 1) + B(π₯ + 3))/(π₯ + 3)(π₯ + 1) Canceling denominator 20π₯+15=A(π₯ + 1) + B(π₯ + 3) Putting π₯=β1 20(β1)+15=A(β1 + 1) + B(β1 + 3) β20+15=AΓ0+B (2) Putting π₯=β1 20(β1)+15=A(β1 + 1) + B(β1 + 3) β20+15=AΓ0+B (2) β5=2B B=(β5)/( 2) Similarly Putting π₯=β3 20(β3)+15=A(β3+1)+B(β3+3) β60+15=A(β2) BΓ0 β45=β2A A=45/2 Hence β«1ββ((5π₯^2)/(π₯^2 + 4π₯ + 3) " " =β«1βγ(5β(20π₯ + 15 )/(π₯^2 + 4π₯ + 3)) ππ₯γ) =β«1βγ5βA/(π₯+3)βγ B/(π₯+1) ππ₯ =β«1βγ5 ππ₯γββ«1βγ(45/2)/(π₯ +_3) ππ₯ββ«1βγ(((β5)/( 2)))/(π₯+1) ππ₯γγ =5π₯β45/2 πππ|π₯+3|+5/2 πππ|π₯+1| Hence F(π₯)=5π₯β5/2 [9 πππ|π₯+3|βπππ|π₯+1|] Now, β«_1^2βγ(5π₯^2)/(π₯^2 + 4π₯ + 3) ππ₯=πΉ(2)βπΉ(1) γ =[5Γ2β5/2 (9 πππ|2+3|βπππ|2+1|)] β [5 Γ1β5/2 (9 πππ|1+3|βπππ|1+1|)] =10β5/2 [9 πππ 5βπππ 3]β5+5/2 [9 πππ 4βπππ 2] =10β5β5/2 [9πππ 5βπππ 3β9πππ 4+πππ 2)] =10β5β5/2 [9πππ 5β9πππ 4β(πππ 3βπππ 2)] =πβπ/π (π π₯π¨π  π/πβπ₯π¨π  π/π) (log a β log b = log π/π)