If x ≠ - 1 and ∫ x 3 + x 2 - x - 1 x 5 + x 4 + 3 x 3 + 3 x 2 + x + 1 tan - 1 x 2 + 1 x d x = A log f x + C , then A - tan f 2 =

Question

If x ≠ - 1 and ∫ x 3 + x 2 - x - 1 x 5 + x 4 + 3 x 3 + 3 x 2 + x + 1 tan - 1 x 2 + 1 x d x = A log f x + C , then A - tan f 2 =, - 3 2, - 1 2, 7 2, - 2

MARKS App · Accepted Answer

∫ x 3 + x 2 − x − 1 ( x 5 + x 4 + 3 x 3 + 3 x 2 + x + 1 ) tan − 1 ( x 2 + 1 x ) d x ∫ ( x + 1 ) ( x 2 − 1 ) ( x + 1 ) ( x 4 + 3 x 2 + 1 ) tan − 1 d x ( x + 1 x ) = ∫ ( 1 − 1 x 2 ) ( ( x + 1 x ) 2 + 1 ) tan - 1 ( x + 1 x ) d x ∴ A = 1 f ( x ) = tan − 1 ( x + 1 x ) A − tan f ( 2 ) = 1 − ( 5 2 ) = − 3 2

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