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Question

# ∫dx(1+√x)√x−x2 is equal to?

A
2(x1)1x+c
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B
2(1x)1x+c
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C
(x1)1x+c
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D
(x+1)1x+c
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Solution

## The correct option is A 2(√x−1)√1−x+cGiven,∫1(√x+1)(√x−x2)dxLet t=√xSo, dt=12√xdx∴dx=2tdtSubstituting,∫1(t+1)(√t2−t4)2tdt=2∫t(t+1)(√t2−t4)dt=2∫dt(t+1)(√t2−t4)substitute t=sinθSo, dt=cosθdθSolving,2∫cosθ dθ(sinθ+1)(√1−(sin2θ)=2∫cosθdθ(√cos2θ)as 1−sin2θ=cos2θ=2∫cosθdθ(sinθ+1)cosθ=2∫dθsinθ+1=2∫1sinθ+1(1−sinθ)(1−sinθ)dθ=2∫1−sinθ1−sin2θdθas (a+b)(a−b)=a2−b2=2∫1−sinθcos2θdθ=2∫1cos2θdθ−2∫sinθcos2θdθ=2∫sec2θdθ−2∫tanθsec2θdθ=2tanθ−2secθ+c=2sinθcosθ−2×1cosθ+c=2sinθ√1−sin2θ−2×1√1−sin2θ+c=2t√1−t2−2×1√1−t2+c=2(t−1)√1−t2+c=2(√x−1)√1−x+cSo, integration of 1(√x−x2)(√x+1)is 2(√x−1)√1−x+c

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