Integration by partial fraction - Type 3

Chapter 7 Class 12 Integrals
Concept wise

### Transcript

Ex 7.5, 11 Integrate the function 5π₯/((π₯ + 1) (π₯2β 4) ) We can write the integrand as 5π₯/((π₯ + 1) (π₯2β 4) ) = 5π₯/((π₯ + 1) (π₯ β 2) (π₯ + 2) ) 5π₯/((π₯ + 1) (π₯2β 4) ) = π΄/((π₯ + 1) ) + π΅/((π₯ β 2) ) + πΆ/((π₯ + 2) ) 5π₯/((π₯ + 1) (π₯2β 4) ) = (π΄(π₯ β 2)(π₯ + 2) + π΅(π₯ + 1)(π₯ + 2) + πΆ(π₯ +1)(π₯ β 2))/((π₯ + 1) (π₯ β 2) (π₯ + 2) ) Cancelling denominator 5π₯ = π΄(π₯β2)(π₯+2)+π΅(π₯+1)(π₯+2)+πΆ(π₯+1)(π₯β2) β¦(1) Putting x = β1 in (1) 5π₯ = π΄(π₯β2)(π₯+2)+π΅(π₯+1)(π₯+2)+πΆ(π₯+1)(π₯β2) 5( β1) = π΄(β1β2)(β1+2)+π΅(β1+1)(β1+2)+πΆ(β1+1)(β1β2) β5 = π΄(β3)(1)+π΅Γ0+πΆΓ0 β5 = β3π΄ π΄ = (β5)/(β3) = 5/3 Putting x = 2 in (1) 5π₯ = π΄(π₯β2)(π₯+2)+π΅(π₯+1)(π₯+2)+πΆ(π₯+1)(π₯β2) 5"(2) = " π΄(2β2)(2+2)+π΅(2+1)(2+2)+πΆ(2+1)(2β2) 10 = π΄Γ0+π΅(3)(4)+πΆΓ0 10 = 12π΅ π΅ = 10/12=5/6 Putting x = β2 in (1) 5π₯ = π΄(π₯β2)(π₯+2)+π΅(π₯+1)(π₯+2)+πΆ(π₯+1)(π₯β2) 5"("β"2) = " π΄(β2β2)(β2+2)+π΅(β2+1)(β2+2)+πΆ(β2+1)(β2β2) β10 = π΄Γ0+π΅Γ0+πΆ(β1)(β4) β10 = 4πΆ πΆ = (β10)/4 πΆ = (β5)/2 Therefore β«1β5π₯/((π₯ + 1) (π₯2β 4) )=β«1β(π΄/(π₯ + 1)+π΅/(π₯ β 2)+πΆ/(π₯ + 2)) ππ₯ =5/3 β«1βππ₯/(π₯ + 1) ππ₯+ 5/6 β«1βππ₯/(π₯ β 2) ππ₯β5/2 β«1βππ₯/((π₯ + 2) ) =π/π γπππ γβ‘|π+π|β π/π γπ₯π¨π  γβ‘|π+π|+π/π γπ₯π¨π  γβ‘|πβπ|+πͺ