If x 1 and (x^3+x^2-x-1)(x^5+x^4+3x^3+3x^2+x+1) tan^-1 ⁡ ( x^2+1x )dx = Alog ⁡ (f (x ) )+C , then A- tan ⁡ (f (2 ) )=

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Correct answer: A. (a) Let I= x^3+x^2-x-1(x^5+x^4+3x^3+3x^2+x+1) tan^-1 ⁡ ( x^2+1x ) dx as x 1 = x(x^2-1)+x^2-1(x(x^4+3x^2+1)+(x^4+3x^2+1)) tan^-1 ⁡ ( x^2+1x ) dx= (x+1 )(x^2-1)(x^4+3x^2+1) (x+1 ) tan^-1 ⁡ ( x^2+1x ) dx= x^2-1(x^4+3x^2+1) tan^-1 ⁡ ( x^2+1x ) dx Let tan^-1 ⁡ ( x^2+1x )=t ( ∵ d dx tan^-1 ⁡ (x )= 11+x^2) ⇒ 11+ ( x^2+1x )^2 2x (x )-(x^2+1)x^2 dx= dt ⇒ x^2-1x^4+3x^2+1 dx= dt ⇒I= 1t dt= log ⁡t+C= log ⁡ tan^-1 ⁡ ( x^2+1x )+C ∴ A=1 ,f (x )= tan^-1 ⁡ ( x^2+1x ) ∴ A- tan ⁡f (2 )=1- tan ⁡ tan^-1 ⁡ 52=1- 52=- 32

$If x \neq = 1 and \sum \frac{( x ^{3} + x ^{2} - x - 1 )}{( x ^{5} + x ^{4} + 3 x ^{3} + 3 x ^{2} + x + 1 ) tan ^{- 1} ( \frac{x ^{2} + 1}{x} )} d x = Alog (f (x)) + C, then A - tan (f (2)) =$

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