1. Chapter 7 Class 12 Integrals
2. Concept wise
3. Integration by partial fraction - Type 5

Transcript

Example 39 Evaluate β«1β(π₯^4 ππ₯)/(π₯ β1)(π₯^2 + 1) Let I = β«1β(π₯^4 ππ₯)/(π₯ β1)(π₯^2 + 1) ππ₯ We can write π₯^4/(π₯ β1)(π₯^2 + 1) = π₯^4/(π₯^3 β π₯^2+ π₯ β 1) Dividing Numerator by denominator as follows. Hence π₯^4 = (π₯^3βπ₯^2+π₯+1) (π₯+1)+1 Thus, π₯^4/(π₯^3 β π₯^2 + π₯ + 1) = (π₯+1)+1/(π₯^3 β π₯^2 + π₯ + 1) = (π₯+1)+1/((π₯ β 1) (π₯^2 +1) ) Now, we can write 1/((π₯^2 + 1) (π₯ β 1) )= (π΄π₯ + π΅)/(π₯^2 + 1) + πΆ/(π₯ β 1) 1/((π₯^2 + 1) (π₯ β 1) )= ((π΄π₯ + π΅)(π₯ β 1) + πΆ (π₯^2 + 1))/((π₯^2 + 1)(π₯ β1)) Canceling denominator 1 = (π΄π₯ + π΅)(π₯ β 1) + πΆ (π₯^2 + 1) Putting x = 1 1 = (π΄(1) + π΅)(1β1) + πΆ ((β1)^2 + 1) 1 = (π΄+π΅)(0)+ πΆ (1+1) 1 = 2πΆ πΆ=1/2 Putting x = 0 1 = (π΄π₯ + π΅)(π₯ β 1) + πΆ (π₯^2 + 1) 1 = (π΄(0) + π΅)(0β1) + πΆ (0^2+1) 1 = (π΅)(β1) + πΆ (1) 1 = πΆ β"B" B =πΆβ1 B =1/2 β1 B =(β1)/2 Putting x = β 1 1 = (π΄π₯ + π΅)(π₯ β 1) + πΆ (π₯^2 + 1) 1 = (π΄(β1)+ π΅)(β1β1) + πΆ ((β1)^2+1) 1 = (βπ΄+π΅)(β2)+πΆ (1+1) 1 = (π΄βπ΅)2+πΆ (2) 1/2=π΄βπ΅+πΆ π΄=1/2+π΅βπΆ π΄ =1/2β1/2β1/2 π΄ =(β1)/2 Hence we can write 1/((π₯^2 + 1) (π₯ β 1) )= (π΄π₯ + π΅)/(π₯^2 + 1) + πΆ/(π₯ β 1) 1/((π₯^2 + 1) (π₯ β 1) ) = (β 1/2 π₯ β 1/2)/(π₯^2 + 1) + (1/2)/(π₯ β 1) = (β1)/2 ( π₯)/(π₯^2 + 1) β1/2 1/(π₯^2 + 1)+ 1/2(π₯ β 1) Hence we can write 1/((π₯^2 + 1) (π₯ β 1) )= (π΄π₯ + π΅)/(π₯^2 + 1) + πΆ/(π₯ β 1) 1/((π₯^2 + 1) (π₯ β 1) ) = (β 1/2 π₯ β 1/2)/(π₯^2 + 1) + (1/2)/(π₯ β 1) = (β1)/2 ( π₯)/(π₯^2 + 1) β1/2 1/(π₯^2 + 1)+ 1/2(π₯ β 1) Therefore, we can write I=β«1βγ(π₯+1)+1/(π₯^2 + 1)(π₯ β 1) ππ₯γ =β«1β[(π₯+1)β1/2 π₯/((π₯^2 + 1) ) ππ₯ββ«1βγ1/2 1/(π₯^2 + 1) ππ₯+β«1βγ1/2 1/((π₯ β 1) ) ππ₯γγ] =π₯^2/2+π₯β1/2 β«1βγπ₯/(π₯^2 + 1)β1/2 β«1βγ1/(π₯^2 + 1) ππ₯+1/2 β«1βγ1/(π₯ β 1) ππ₯γγγ β΄ I = π₯^2/2+π₯ β 1/2 I"1 β " 1/2 I"2 + " 1/2 I"3" Solving π°π I1=β«1βγπ₯/(π₯^2 + 1) ππ₯γ Put π‘=π₯^2+1 Differentiating w.r.t. π₯ ππ‘/ππ₯=2π₯+0 ππ‘/2π₯=ππ₯ Therefore, β«1βγ(π₯ ππ₯)/(π₯^2 + 1)=β«1βπ₯/π‘ ππ‘/2π₯γ=β«1β1/2 ππ‘/π‘=1/2 πππ|π‘|+πΆ1 Putting π‘=π₯^2+1 =1/2 πππ|π₯^2+1|+πΆ1 And, I2=β«1βγ1/(π₯^2 + 1) ππ₯γ=tan^(β1)β‘γπ₯+γ πΆ2 I3=β«1βγ1/(π₯ β1) ππ₯γ=πππ|π₯β1|+πΆ3 Hence πΌ=π₯^2/2+π₯β1/2 πΌ1β1/2 πΌ2+1/2 πΌ3 =π₯^2/2+π₯β1/2 (1/2 πππ|π₯^2+1|+πΆ1)β1/2 (γπ‘ππγ^(β1) (π₯)+C_2 )β1/2 (πππ|π₯β1|+πΆ3) =π₯^2/2+π₯β1/4 πππ|π₯^2+1|+πΆ1/2β1/2 tan^(β1)β‘γπ₯ πΆ2/2+1/2 πππ|π₯β1|+πΆ3/2γ =π^π/π+π+π/π πππ|πβπ|βπ/π πππ(π^π+π)βπ/π γπππγ^(βπ)β‘γπ+πͺγ

Integration by partial fraction - Type 5