Integration Full Chapter Explained - https://you.tube/Integration-Class-12

Last updated at Dec. 23, 2019 by Teachoo

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Misc 41 β«βππ₯/(π^π₯ + π^(βπ₯) ) is equal to (A) tan^(β1) (π^π₯ )+πΆ (B) tan^(β1)β‘γ(π^(βπ₯) )+πΆγ (C) logβ‘(π^π₯βπ^(βπ₯) )+πΆ (D) logβ‘(π^π₯+π^(βπ₯) )+πΆ β«βππ₯/(π^π₯ + π^(βπ₯) ) = β«βππ₯/(π^π₯ + 1/π^π₯ ) = β«1β(π^π₯ ππ₯)/(π^2π₯ + 1) Let π^π₯=π‘ ππ‘/ππ₯=π^π₯ dt = π^π₯ ππ₯ Substituting, = β«1βππ‘/(π‘^2 +1) = γπ‘ππγ^(β1) (π‘)+ C Putting value of t = γπππγ^(βπ) (π^π )+ C Hence, answer is (A). (β«1βγππ₯/(π₯^2 + 1)=γπ‘ππγ^(β1) π₯γ " " )

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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.