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Question

If dx(xα)(βx)=Atan1f(x)+C then

A
A=1/2
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B
f(x)=⎢ ⎢ ⎢ ⎢x(α+β2)(xα)(βx)⎥ ⎥ ⎥ ⎥
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C
f(x)=(xα)(βx)
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D
A=1
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Solution

The correct options are
A f(x)=⎢ ⎢ ⎢ ⎢x(α+β2)(xα)(βx)⎥ ⎥ ⎥ ⎥
D A=1
Given : dx(xα)(βx)=Atan1f(x)+C
I=dx(xα)(βx)=dx(α+β)4αβ[x(α+β2)]2

I=dx(α+β)4[x(α+β2)]2=sin1⎢ ⎢ ⎢ ⎢x(α+β2)αβ2⎥ ⎥ ⎥ ⎥
$

I=tan1x(α+β2)(α+β)4αβ[x(α+β2)]2

I=tan1⎢ ⎢ ⎢ ⎢x(α+β2)x2+(α+β)xαβ⎥ ⎥ ⎥ ⎥=tan1⎢ ⎢ ⎢ ⎢x(α+β2)(xα)(βx)⎥ ⎥ ⎥ ⎥+C

A=1f(x)=⎢ ⎢ ⎢ ⎢x(α+β2)(xα)(βx)⎥ ⎥ ⎥ ⎥


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