The value of `int_(0)^(1)(x^(a)-1)/(logx)dx` is

Options

A. `log(a-1)`
B. `log(a+1)`
C. `alog(a+1)`
D. none of these

Answer

B

Explanation

Let `I(a)=int_(0)^(1)(x^(a)-1)/(logx)dx`……………..1
Differentiating w.r.t `a` keeping `x` is constant, we get
`(dI(a))/(da)=int_(0)^(1)((x^(a)-1)/(logx))dx`
`=int_(0)^(1)(x^(a)logx)/(logx)dx`
`=int_(0)^(1)x^(a)dx`
`=(x^(a+1))/(a+1)|_(0)^(1)=1/((a+1))`
Integrating both sides w.r.t `a` we get
`I(a)=log(a+1)+c`
For `a=0,1(0)=log1+c` [from equation 1]
`0=0+c`
`:. I=log(a+1)`