∫ 0 1 x tan - 1 ⁡ x d x =

Question

∫ 0 1 x tan - 1 ⁡ x d x =, π 4 + 1 2, π 4 - 1 2, 1 2 - π 4, - π 4 - 1 2

MARKS App · Accepted Answer

∫ 0 ( u v r u l e ) 1 x tan - 1 ⁡ x d x = tan - 1 ⁡ x ∫ x d x 0 1 - ∫ 0 1 d d x tan - 1 ⁡ x ∫ x d x d x = tan - 1 ⁡ x ⋅ x 2 2 0 1 - ∫ 0 1 1 1 + x 2 ⋅ x 2 2 d x = π 4 ⋅ 1 2 - 0 - 1 2 ∫ 0 1 x 2 1 + x 2 d x = π 8 - 1 2 ∫ 0 1 1 - 1 1 + x 2 d x = π 8 - 1 2 x - tan - 1 ⁡ x 0 1 = π 8 - 1 2 1 - 0 - π 4 - 0 = π 8 - 1 2 + π 8 = π 4 - 1 2

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