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Ex 7.5
Ex 7.5, 1
Ex 7.5, 2
Ex 7.5, 3 Important
Ex 7.5, 4
Ex 7.5, 5
Ex 7.5, 6 Important
Ex 7.5, 7 Important
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Ex 7.5, 9 Important
Ex 7.5, 10
Ex 7.5, 11 Important
Ex 7.5, 12
Ex 7.5, 13 Important
Ex 7.5, 14 Important
Ex 7.5, 15
Ex 7.5, 16 Important
Ex 7.5, 17
Ex 7.5, 18 Important
Ex 7.5, 19
Ex 7.5, 20 Important You are here
Ex 7.5, 21 Important
Ex 7.5, 22 (MCQ)
Ex 7.5, 23 (MCQ) Important
Ex 7.6β
Chapter 7 Class 12 Integrals
Serial order wise
- Ex 7.1
- Ex 7.2
- Ex 7.3
- Ex 7.4
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Ex 7.5
- Ex 7.6
- Ex 7.7
- Ex 7.8
- Ex 7.9
- Ex 7.10
- Examples
- Miscellaneous
- Case Based Questions (MCQ)
- NCERT Exemplar MCQ
- Area as a sum
Last updated at April 16, 2024 by Teachoo
Transcript
Ex 7.5, 20Integrate the function 1/(π₯(π₯4β1) ) 1/(π₯(π₯4 β 1) ) Multiplying integrand by π₯^3/π₯^3 = 1/(π₯(π₯^4 β 1) ) Γ π₯^3/π₯^3 = π₯^3/(π₯^4 (π₯^4 β 1) ) Let t = π₯^4Differentiating both sides π€.π.π‘.π₯ ππ‘/ππ₯ = 4π₯^3 ππ‘/(4π₯^3 ) = ππ₯Substituting value of π‘ = π₯^4 & ππ₯ = ππ‘/(4π₯^3 ) " "β«1βπ₯^3/(π₯^4 (π₯^4β 1) ) ππ₯ = β«1βπ₯^3/(π‘(π‘ β 1) ) ππ‘/(4π₯^3 ) " " = 1/4 β«1βππ‘/(π‘(π‘ β 1) ) We can write integrand as 1/(π‘(π‘ β 1) ) = π΄/π‘ + π΅/(π‘ β 1) 1/(π‘(π‘ β 1) ) = (π΄(π‘ β 1) + π΅ π‘)/π‘(π‘ β 1) Cancelling denominator 1 = π΄(π‘β1)+π΅π‘β¦(1)Putting t = 0 in (1) 1 = π΄(0β1)+π΅Γ0 1 = π΄Γ(β1) 1 = βπ΄ π΄ = β1Putting t = 1 in (1) 1 = A(tβ1)+Bt 1 = π΄(1β1)+π΅Γ1 1 = π΄Γ0+π΅ 1 = π΅ π΅ = 1Therefore 1/4 β«1β1/(π‘(π‘ β 1) ) ππ‘ = β«1β(β1)/(π‘ ) ππ‘ + β«1β1/(π‘ β 1 ) = βγlog γβ‘|π‘|+γlog γβ‘|π‘β1|+πΆ = γlog γβ‘|(π‘ β 1)/π‘|+πΆ Putting back t =γ π₯γ^4 = π/π γπ₯π¨π γβ‘|(π^π β π)/π^π |+πͺ ("As " πππ π΄βπππ π΅" = " πππ π΄/π΅)
Next: Ex 7.5, 21 Important β Go Ad-free
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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.
