If x∈(0,1) and ∫xtan(2tan−1(1+x+1−1+x−11+x+1+1+x−1)) dx=AxB+C, then (A, B are positive constants and C is constant of integration)

1+x=sec4θ tan−1(sec2θ+1−sec2θ−1sec2θ+1+sec2θ−1)=tan−1(1+cos2θ−1−cos2θ1+cos2θ+1−cos2θ) =tan−1(2.cosθ−2sinθ2cosθ+2sinθ)=tan−1(cosθ−sinθcosθ+sinθ)=tan−1(tan(π4−θ)) =π4−θ ∫x.tan(2(π4−θ)) dx=∫x.(cot2θ)dx=∫xxdx =∫x14dx=45x54+C A=45,B=54⇒AB=1

If x∈(0,1) and ∫xtan(2tan−1(1+x+1−1+x−11+x+1+1+x−1)) dx=AxB+C, then (A, B are positive constants and C is constant of integration)

1+x=sec4θ tan−1(sec2θ+1−sec2θ−1sec2θ+1+sec2θ−1)=tan−1(1+cos2θ−1−cos2θ1+cos2θ+1−cos2θ) =tan−1(2.cosθ−2sinθ2cosθ+2sinθ)=tan−1(cosθ−sinθcosθ+sinθ)=tan−1(tan(π4−θ)) =π4−θ ∫x.tan(2(π4−θ)) dx=∫x.(cot2θ)dx=∫xxdx =∫x14dx=45x54+C A=45,B=54⇒AB=1

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If $x \in (0, 1)$ and $\int \sqrt{x} \tan (2 \tan^{- 1} (\frac{\sqrt{\sqrt{1 + \sqrt{x}} + 1} - \sqrt{\sqrt{1 + \sqrt{x}} - 1}}{\sqrt{\sqrt{1 + \sqrt{x}} + 1} + \sqrt{\sqrt{1 + \sqrt{x}} - 1}})) d x = A x^{B} + C$ , then (A, B are positive constants and C is constant of integration)

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$A + B = \frac{21}{20}$

$A B = 1$

$A + B = \frac{13}{6}$

$A - B = 1$

answer is C.

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Detailed Solution

$\sqrt{1 + \sqrt{x}} = \sec 4 θ \tan^{- 1} (\frac{\sqrt{\sec 2 θ + 1} - \sqrt{\sec 2 θ - 1}}{\sqrt{\sec 2 θ + 1} + \sqrt{\sec 2 θ - 1}}) = \tan^{- 1} (\frac{\sqrt{1 + \cos 2 θ} - \sqrt{1 - \cos 2 θ}}{\sqrt{1 + \cos 2 θ} + \sqrt{1 - \cos 2 θ}}) = \tan^{- 1} (\frac{\sqrt{2} . \cos θ - \sqrt{2} \sin θ}{\sqrt{2} \cos θ + \sqrt{2} \sin θ}) = \tan^{- 1} (\frac{\cos θ - \sin θ}{\cos θ + \sin θ}) = \tan^{- 1} (\tan (\frac{π}{4} - θ)) = \frac{π}{4} - θ \int \sqrt{x} . \tan (2 (\frac{π}{4} - θ)) d x = \int \sqrt{x} . (\cot 2 θ) d x = \int \frac{\sqrt{x}}{\sqrt{\sqrt{x}}} d x = \int x^{\frac{1}{4}} d x = \frac{4}{5} x^{\frac{5}{4}} + C A = \frac{4}{5}, B = \frac{5}{4} \Rightarrow A B = 1$