Integration Full Chapter Explained - Integration Class 12 - Everything you need
Example 39 - Chapter 7 Class 12 Integrals
Last updated at Dec. 23, 2019 by Teachoo
Transcript
Example 39 Evaluate β«1β(π₯^4 ππ₯)/(π₯ β1)(π₯^2 + 1) Let I = β«1β(π₯^4 ππ₯)/(π₯ β1)(π₯^2 + 1) ππ₯ We can write π₯^4/(π₯ β1)(π₯^2 + 1) = π₯^4/(π₯^3 β π₯^2+ π₯ β 1) Dividing Numerator by denominator as follows. Hence π₯^4 = (π₯^3βπ₯^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) + πΆ/(π₯ β 1) 1/((π₯^2 + 1) (π₯ β 1) )= ((π΄π₯ + π΅)(π₯ β 1) + πΆ (π₯^2 + 1))/((π₯^2 + 1)(π₯ β1)) Canceling denominator 1 = (π΄π₯ + π΅)(π₯ β 1) + πΆ (π₯^2 + 1) Putting x = 1 1 = (π΄(1) + π΅)(1β1) + πΆ ((β1)^2 + 1) 1 = (π΄+π΅)(0)+ πΆ (1+1) 1 = 2πΆ πΆ=1/2 Putting x = 0 1 = (π΄π₯ + π΅)(π₯ β 1) + πΆ (π₯^2 + 1) 1 = (π΄(0) + π΅)(0β1) + πΆ (0^2+1) 1 = (π΅)(β1) + πΆ (1) 1 = πΆ β"B" B =πΆβ1 B =1/2 β1 B =(β1)/2 Putting x = β 1 1 = (π΄π₯ + π΅)(π₯ β 1) + πΆ (π₯^2 + 1) 1 = (π΄(β1)+ π΅)(β1β1) + πΆ ((β1)^2+1) 1 = (βπ΄+π΅)(β2)+πΆ (1+1) 1 = (π΄βπ΅)2+πΆ (2) 1/2=π΄βπ΅+πΆ π΄=1/2+π΅βπΆ π΄ =1/2β1/2β1/2 π΄ =(β1)/2 Hence we can write 1/((π₯^2 + 1) (π₯ β 1) )= (π΄π₯ + π΅)/(π₯^2 + 1) + πΆ/(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) = (β1)/2 ( π₯)/(π₯^2 + 1) β1/2 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) + πΆ/(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) = (β1)/2 ( π₯)/(π₯^2 + 1) β1/2 1/(π₯^2 + 1)+ 1/2(π₯ β 1) Therefore, we can write I=β«1βγ(π₯+1)+1/(π₯^2 + 1)(π₯ β 1) ππ₯γ =β«1β[(π₯+1)β1/2 π₯/((π₯^2 + 1) ) ππ₯ββ«1βγ1/2 1/(π₯^2 + 1) ππ₯+β«1βγ1/2 1/((π₯ β 1) ) ππ₯γγ] =π₯^2/2+π₯β1/2 β«1βγπ₯/(π₯^2 + 1)β1/2 β«1βγ1/(π₯^2 + 1) ππ₯+1/2 β«1βγ1/(π₯ β 1) ππ₯γγγ 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«1βγπ₯/(π₯^2 + 1) ππ₯γ Put π‘=π₯^2+1 Differentiating w.r.t. π₯ ππ‘/ππ₯=2π₯+0 ππ‘/2π₯=ππ₯ Therefore, β«1βγ(π₯ ππ₯)/(π₯^2 + 1)=β«1βπ₯/π‘ ππ‘/2π₯γ=β«1β1/2 ππ‘/π‘=1/2 πππ|π‘|+πΆ1 Putting π‘=π₯^2+1 =1/2 πππ|π₯^2+1|+πΆ1 And, I2=β«1βγ1/(π₯^2 + 1) ππ₯γ=tan^(β1)β‘γπ₯+γ πΆ2 I3=β«1βγ1/(π₯ β1) ππ₯γ=πππ|π₯β1|+πΆ3 Hence πΌ=π₯^2/2+π₯β1/2 πΌ1β1/2 πΌ2+1/2 πΌ3 =π₯^2/2+π₯β1/2 (1/2 πππ|π₯^2+1|+πΆ1)β1/2 (γπ‘ππγ^(β1) (π₯)+C_2 )β1/2 (πππ|π₯β1|+πΆ3) =π₯^2/2+π₯β1/4 πππ|π₯^2+1|+πΆ1/2β1/2 tan^(β1)β‘γπ₯ πΆ2/2+1/2 πππ|π₯β1|+πΆ3/2γ =π^π/π+π+π/π πππ|πβπ|βπ/π πππ(π^π+π)βπ/π γπππγ^(βπ)β‘γπ+πͺγ
Examples
Example 1
Example 2
Example 3 Important
Example 4
Example 5
Example 6 Important
Example 7 Important
Example 8
Example 9
Example 10 Important
Example 11
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20 Important
Example 21 Important
Example 22 Important
Example 23
Example 24
Example 25 Important Not in Syllabus - CBSE Exams 2021
Example 26 Not in Syllabus - CBSE Exams 2021
Example 27 Important
Example 28
Example 29
Example 30 Important
Example 31
Example 32
Example 33
Example 34 Important
Example 35 Important
Example 36 Important
Example 37
Example 38 Important
Example 39 Important You are here
Example 40 Important
Example 41 Important
Example 42 Important
Example 43 Important
Example 44 Important
Example 25 (Supplementary NCERT) Important Not in Syllabus - CBSE Exams 2021
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.