If ∫ log t + 1 + t 2 1 + t 2 d t = 1 2 g t 2 + c , where c is a constant, then g 2 , is equal to

If ∫logt+1+t21+t2dt=12gt2+c, where c is a constant, then g2, is equal to
  1. 15log2+5
  2. 2log2+5
  3. log2+5
  4. 12log(2+5)

Solution

Put, logt+1+t2=z

⇒1t+1+t21+2t21+t2dt=dz

⇒t+1+t2 t+1+t2 11+t2dt=dz

⇒11+t2dt=dz

∴I=∫zdz=z22+c=12logt+1+t22+c, where c is the constant of integration.

∴gt=logt+1+t2

⇒g2=log2+ 1+4

Hence, log2+5, is the answer.

Asked in: JEE Main 2015 (11 Apr Online)

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