Integration by partial fraction - Type 3

Chapter 7 Class 12 Integrals
Concept wise

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Misc 1 (Method 1) Integrate the function 1/(π₯ β π₯^3 ) Solving integrand 1/(π₯ β π₯^3 )=1/π₯(1 β π₯^2 ) =π/π(π β π)(π + π) We can write it as π/π(π β π)(π + π) = π¨/π + π©/((π β π) ) + π/((π + π) ) 1/π₯(1 β π₯)(1 + π₯) = (π΄(1 β π₯) (1 + π₯) + π΅π₯ (1 + π₯) + πΆπ₯ (1 β π₯))/( π₯ (1 β π₯) (1 + π₯) ) Cancelling denominator π = π¨(π β π) (π + π) + π©π (π + π) + πͺπ (π β π) Putting π=π in (1) 1=π΄(1 β 0) (1 + 0) + π΅(0) (1 + 0) + πΆ(0) (1 β 0) 1=π΄ Γ 1 Γ 1 + π΅ Γ 0 + πΆ Γ 0 1=π΄+0+0 π¨=π Putting π=π in (1) 1 = π΄(1 β1) (1 +1) + π΅(1) (1 +1) + πΆ(1) (1 β1) 1 = π΄ Γ 0 + π΅ Γ (1) Γ (2) + πΆ Γ 0 1 = 0 +2π΅ + 0 π© = π/π Putting π=βπ in (1) 1 = π΄(1 β(β1)) (1 +(β1)) + π΅(β1) (1 +(β1)) + πΆ(β1) (1 β(β1)) 1 = π΄(1 + 1) (1 β1) + π΅(β1)(1β1) + πΆ(β1)(1+1) 1 = π΄ Γ 0 + π΅ Γ 0 + πΆ Γ(β1)(2) 1 = 0 +0 β2πΆ 1 = β2πΆ πͺ = βπ/π Hence we can write it as 1/π₯(1 β π₯)(1 + π₯) = 1/π₯ + (π/π)/((1 β π₯) ) + (β1/2)/((1 + π₯) ) π/π(π β π)(π + π) = π/π + π/π(π β π) + (βπ)/π(π + π) Therefore β«1βπ/π(π β π)(π + π) ππ = β«1β1/π₯ ππ₯ + β«1β1/2(1 β π₯) ππ₯ + β«1β(β1)/2(1 + π₯) ππ₯ = β«1β1/π₯ ππ₯ + 1/2 β«1β1/((1 β π₯) ) ππ₯ β 1/2 β«1β1/((1 + π₯) ) ππ₯ = γπ₯π¨π  γβ‘|π|+π/π [γπ₯π¨π  γβ‘|π β π|/(βπ)] β1/2 γπ₯π¨π  γβ‘|π + π|+πͺ = γlog γβ‘|π₯|β γ 1/2 log γβ‘|1 β π₯|β1/2 γlog γβ‘|1 + π₯|+πΆ = γlog γβ‘|π₯|β1/2 [γlog γβ‘|1 β π₯|+γlog γβ‘|1 + π₯| ]+πΆ = γlog γβ‘|π₯|β1/2 [γlog γβ‘|1 β π₯| |1 + π₯|]+πΆ As πππ π¨+πππ π©=logβ‘π΄π΅ = γlog γβ‘|π₯|β1/2 [γlog γβ‘|(1 β π₯^2 )| ]+πΆ = 1/2 [2 γlog γβ‘|π₯|βγlog γβ‘|(1 β π₯^2 )|+2πΆ] = 1/2 [γlog γβ‘γ|π₯|^2 γβγlog γβ‘|(1 β π₯^2 )|+πΎ] = π/π log |π^π/((π β π^π ) )|+π² As πππ π¨βπππ π©=πππ π΄/π΅ Misc 1 (Method 2) Integrate the function 1/(π₯ β π₯^3 ) Now, β«1βγ1/(π₯ β π₯^3 ) ππ₯γ Taking x3 common from the denominator =β«1β1/(π₯^3 (π₯/π₯^3 β 1) ) ππ₯ =β«1βπ/(π^π (π/π^π β π) ) ππ Let t = π/π^π βπ Differentiating with respect to π₯ π/ππ₯ (1/π₯^2 β1)=ππ‘/ππ₯ (β2)/π₯^3 =ππ‘/ππ₯ ππ=(π^π ππ)/(βπ) Putting the value t and dt in the equation β«1βγπ/(π^π (π/π^π βπ) ) ππγ=β«1βγ1/(π₯^3 (π‘) ) Γ (π₯^3 ππ‘)/(β2)γ =β«1βγπ/(βπ) ππ/πγ =(β1)/( 2) β«1βππ‘/π‘ =(β1)/( 2) πππ|π‘|+πΆ Putting back π=π/π^π βπ =(β1)/( 2) πππ|1/π₯^2 β1|+πΆ =(β1)/( 2) πππ|(π β π^π)/π^π |+πΆ = 1/2 log |(1 β π₯^2)/π₯^2 |^(βπ)+πΆ = π/π log |π^π/(π β π^π )|+πͺ (As a log b = log π^π)