2
Question
Solve : ∫βαdx(x−α)(β−x),β>α
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Solution
1(x−α)(β−x)=1β−α((β−x)+(x−α)(x−α)(β−x))=1β−α((β−x)(x−α)(β−x))+1β−α((x−α)(x−α)(β−x))=1β−α1x−α+1β−α1β−xIntegrating w.r.t x we get=1β−α∫dxx−α+1β−α∫dxβ−x=1β−αln|x−α|−1β−αln|β−x|=1β−α[ln|x−α|−ln|β−x|]βα=1β−α[ln|β−α|−ln|α−α|−ln|β−β|+ln|β−α|]
=1β−α[ln|β−α|−1−1+ln|β−α|]=2β−α[ln|β−α|−1]
1(x−α)(β−x)
=1β−α((β−x)+(x−α)(x−α)(β−x))
=1β−α((β−x)(x−α)(β−x))+1β−α((x−α)(x−α)(β−x))
=1β−α1x−α+1β−α1β−x
Integrating w.r.t x we get
=1β−α∫dxx−α+1β−α∫dxβ−x
=1β−αln|x−α|−1β−αln|β−x|
=1β−α[ln|x−α|−ln|β−x|]βα
=1β−α[ln|β−α|−ln|α−α|−ln|β−β|+ln|β−α|]
=1β−α[ln|β−α|−1−1+ln|β−α|]
=2β−α[ln|β−α|−1]
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