Ex 7.2, 38 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Aug. 9, 2021 by
Last updated at Aug. 9, 2021 by
Transcript
Ex 7.2, 38 β«1β(γ10π₯γ^9+γ10γ^π₯ log_πβ‘10)/(π₯10+ 10π₯) dx equals (A) 10π₯ β π₯^10 + πΆ (B) 10π₯+π₯^10+πΆ (C) (10π₯ β π₯^10 )^(β1) + πΆ (D) logβ‘(10π₯+π₯10) + πΆ Let π₯10+ 10π₯= π‘ Differentiating both sides π€.π.π‘.π₯ γ10π₯γ^(10β1)+γ10γ^π₯ πππβ‘10= ππ‘/ππ₯ γ10π₯γ^9+γ10γ^π₯ πππβ‘10= ππ‘/ππ₯ ππ₯= ππ‘/(γ10π₯γ^9 + γ10γ^π₯ πππβ‘10 ) (Using (π^π₯ )^β²=π^π₯ πππβ‘π) Now, our function becomes β«1βγ" " (10π₯γ9+10γ^π₯ πππβ‘10)/(π₯^10 + γ10γ^π₯ )γ . ππ₯ Putting (π₯^10+ γ10γ^π₯ )=π‘ & ππ₯=" " ππ‘/(γ10π₯γ^9 + γ10γ^π₯ πππβ‘10 ) = β«1βγ" " (10π₯γ9+10γ^π₯ πππβ‘10)/π‘γ . ππ‘/(γ10π₯γ^9+γ10γ^π₯ πππβ‘10 ) " " = β«1βγ" " 1/π‘γ.ππ‘ = log |π‘|+πΆ = log |γ10γ^π₯+ π₯^10 |+πΆ = log (γ10γ^π₯+ π₯^10 )+πΆ β΄ Option D is correct. (Using π‘=γ10γ^π₯+π₯^10) (As 10x and x10 are positive)
Integration by substitution - lnx
Integration by substitution - lnx
About the Author