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Kindly consider the following

0 2 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Long answer type Questions as classified in NCERT Exemplar

    Sol:

    Let  I=∫01xlogII|1+2x|Idx              =[log|1+2x|.(x22)]01−∫01(1.21+2x.x22)dx              =12[x2log(1+2x)]01−∫01(x21+2x)dx              =12[log3−0]−∫01(x2−x/21+2x)dx              =12log3−12∫01x dx+12∫01x1+2xdx              =12log3−12[x22]01+12.12∫01(2x+1−1)1+2xdx              =12log3−14[1−0]+14∫011 dx−14∫0112x+1 dx              =12log3−14+14[x]01−14.12[log|2x+1|]01              =12log3−14+14−18[log3−0]              =12log3−18log3=38log3Hence,  I=38log3.

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