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Kindly consider the following

∫2x−12x+3 dx=x−log∣(2x+3)2∣+C

0 3 Views | Posted 3 months ago
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  • 1 Answer

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    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Short Answer Type Questions as classified in NCERT Exemplar

    Sol:

          L.H.S.=∫2x−12x+3 dx    ⇒∫(1−42x+3) dx      [Dividing  the  numerator  by  the denominator ]    ⇒∫1. dx−4∫12x+3 dx     ⇒∫1. dx−42∫1x+32 dx    ⇒∫1. dx−2∫1x+32 dx      ⇒x−2log|x+32|+C    ⇒x−2log|2x+32|+C      ⇒x−log|(2x+32)2|+C                                                                                     [?nlogm=logmn]    ⇒x−log|(2x+3)2|−log22+C    ⇒x−log|(2x+3)2|+C1=R.H.S.         [where  C1=C−log22]L.H.S.=R.H.S.Hence,  proved.

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