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Question

The value of 1+xxdx is
  1. ln1+x11+x+1+c
  2. 21+x+c
  3. 1+x11+x+1+c
  4. 21+x+ln1+x11+x+1+c


Solution

The correct option is D 21+x+ln1+x11+x+1+c
I=1+xxdx,1+x=t2,dx=2t dtI=(t)(2t) dtt21=2t21+1t21dt=2dt+2(1t11t+1)dt=2t+ln(t1t+1)+c=21+x+ln(1+x11+x+1)+c

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