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Evaluate: int1/((x-alpha)(beta-x))dx ,(b...

Evaluate: `int1/((x-alpha)(beta-x))dx ,(beta>alpha)`

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We have given
`int1/((x-alpha)(beta-x))dx`
`x=alphacos^2theta+betacos^2theta`
`x−alpha=alphasin^2theta+betacos^2theta−alpha`
`=(beta−alpha)cos^2theta`
`beta−x=beta−alphasin^2theta−betacos^2theta`
`=(beta−alpha)sin^2alpha`
`dx=[alpha.2sinthetacostheta+beta2costheta(−sintheta)]`
...
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