Ex 7.5

Chapter 7 Class 12 Integrals
Serial order 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

Made by

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.