Definite Integration - By Partial Fraction
Definite Integration - By Partial Fraction
Last updated at April 16, 2024 by Teachoo
Ex 7.8, 16 β«_1^2β(5π₯^2)/(π₯^2 + 4π₯ + 3) ππ₯ Let F(π₯)=β«1βγ(5π₯^2)/(π₯^2 + 4π₯ + 3) ππ₯γ =5β«1βγπ^π/(π₯^2 + 4π₯ + 3) ππ₯γ =5β«1βγ(π^π + ππ + π β ππ β π)/(π^π + ππ + π) ππ₯γ =5β«1βγ(π₯^2 + 4π₯ + 3)/(π₯^2 + 4π₯ + 3) ππ₯γβ5β«1βγ( (4π₯ + 3))/(π₯^2 + 4π₯ + 3) ππ₯γ =β«1βγ(πβ(πππ + ππ )/(π^π + ππ + π)) π πγ =β«1βγ(5β(20π₯ + 15 )/(π₯^2 + 3π₯ + π₯ + 3)) ππ₯γ =β«1βγ(5β(20π₯ + 15 )/(π₯(π₯ + 3) + 1(π₯ + 3))) ππ₯γ =β«1βγ(πβ(πππ + ππ )/((π + π) (π + π))) π πγ Now, Let (πππ + ππ)/(π + π)(π + π) =π/(π + π)+π/(π + π) (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=βπ 20(β1)+15=A(β1 + 1) + B(β1 + 3) β20+15=AΓ0+B (2) β5=2B π=(βπ)/( π) Putting π=βπ 20(β3)+15=A(β3+1)+B(β3+3) β60+15=A(β2) BΓ0 β45=β2A π¨=ππ/π Hence β«1ββ((ππ^π)/(π^π + ππ + π) " " =β«1βγ(5β(20π₯ + 15 )/(π₯^2 + 4π₯ + 3)) ππ₯γ) =β«1βγ5βA/(π₯ + 3)βγ B/(π₯ + 1) ππ₯ =β«1βγπ π πγββ«1βγ(ππ/π)/(π + π) π πββ«1βγ(((βπ)/( π)))/(π + π) π πγγ =5π₯β45/2 πππ|π₯+3|+5/2 πππ|π₯+1| Hence F(π)=ππβπ/π [π πππ|π+π|βπππ|π+π|] Now, β«_1^2βγ(ππ^π)/(π^π + ππ + π) π π=πΉ(2)βπΉ(1) γ =[5 Γ 2β5/2 (9 πππ|2+3|βπππ|2+1|)] β [5 Γ 1β5/2 (9 πππ|1+3|βπππ|1+1|)] =10β5/2 [9 πππ 5βπππ 3]β5+5/2 [9 πππ 4βπππ 2] =10β5β5/2 [9πππ 5βπππ 3β9πππ 4+πππ 2)] =10β5β5/2 [9πππ 5β9πππ 4β(πππ 3βπππ 2)] =πβπ/π (π π₯π¨π π/πβπ₯π¨π π/π)