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Ex 7.5

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Ex 7.5, 22 (MCQ)

Ex 7.5, 23 (MCQ) Important You are here

Last updated at May 29, 2023 by Teachoo

Ex 7.5, 23 β«1βππ₯/(π₯(π₯2 + 1) ) equals log |π₯| β 1/(2 ) log (x2+1) + C log |π₯| + 1/(2 ) log (x2+1) + C log |π₯| + 1/(2 ) log (x2+1) + C 1/(2 ) log |π₯| + log (x2+1) + C β«1βππ₯/(π₯(π₯2 + 1) ) Let t = (1+π₯^2) Differentiating both sides π€.π.π‘.π₯ ππ‘/ππ₯ = 0+2π₯ ππ₯ = ππ‘/2π₯ Thus, our equation becomes β«1βππ₯/(π₯(π₯^2 + 1) ) = β«1β1/(π₯ π‘) ππ‘/2π₯ = 1/2 β«1βγ ππ‘/π‘Γ1/π₯^2 γ = 1/2 β«1βππ‘/(π‘(π‘ β 1)) We can write integrand as Since t = (1+π₯^2) x2 = (t β 1) 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‘ Putting t = 0 in (1) 1 = π΄(0β1)+π΅Γ0 1 = βπ΄ π΄ = β1 Putting t = 1 in (1) 1 = π΄(1β1)+π΅(1) 1 = π΄Γ0+π΅ 1 = π΅ π΅ = 1 Therefore 1/2 β«1βππ‘/(π‘(π‘ β 1)) = 1/2 β«1βγ(β1)/π‘ γ ππ‘ + 1/2 β«1βγ1/(π‘ β 1) γ ππ‘ = (β1)/2 γlog γβ‘|π‘|+ 1/2 γlog γβ‘|π‘β1|+πΆ Putting back t = (1+π₯^2) = (β1)/2 γlog γβ‘|1+π₯^2 |+ 1/2 γlog γβ‘|1+π₯^2β1|+πΆ = (β1)/2 γlog γβ‘|1+π₯^2 |+ 1/2 γlog γβ‘|π₯^2 |+πΆ = (β1)/2 γlog γβ‘|1+π₯^2 |+ 1/2Γ2 γlog γβ‘|π₯|+πΆ = (β1)/2 γlog γβ‘|1+π₯^2 |+ γlog γβ‘|π₯|+πΆ As 1 + π₯^2 it is always positive = γlog γβ‘|π₯|β 1/2 γlog γβ‘γ(π₯^2+1)γ+πΆ β΄ Correct answer is A . (log π₯^π=π logβ‘π₯)