If α =   ∫ 0 1 e 9 x + 3 tan - 1 ⁡ x   12 + 9 x 2 1 + x 2   d x where tan -…

If α= 01e9x+3tan-1x 12+9x21+x2 dx where tan-1x  takes only principal values, then the value of loge1+α- 3π4 is

Solution

α= 01e9x+3tan-1x 12+9x21+x2 dx
= 01e9x+3tan-1x 9+31+x2 dx
= e9x+3tan-1x|01
= e9+3π4-1
  ln (1+α)- 3π4=ln1+e9+3π4-1-3π4=lne9+3π4-3π4=9 .

Asked in: JEE Advanced 2015 (Paper 2)

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