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integrate int0^1 (x^alpha - 1)/log x...

integrate ` int_0^1 (x^alpha - 1)/log x `

Text Solution

Verified by Experts

Let `I(alpha) = int_(0)^1 (x^alpha -1)/logx` Differentiating it we get,
`I'(alpha) = int_(0)^1 x^(alpha) dx = 1/(alpha+1)`
`:. I(alpha) = log(1+alpha)+c`
If we put `alphaa = 0`,
`0 = log1+c=> c = 0` (As log1 = 0)
`:. I(alpha) = log(1+alpha)`
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