Answer 1

let `int_0^1( x^a-1)/log x dx`  

Taking ‘a’ as parameter , 

I (a)= `int_0^1(x^a-1)/log x dx`        -------- (1) 

differentiate w.r.t a , 

`(dI(a))/(da)=d/(da) int_0^1 (x^a-1)/log x dx` 

∴`(dI(a))/(da)=int_0^1 del/del_a (x^a-1)/log x dx ………{ D.U.I.S f(x)}`  

∴`(dI(a))/(da)= int_0^1 (x^a log x)/log x    dx ……… {(dx^a)/(da)=x^a. log a}` v

∴`( dI(a))/(da)= int_0^1 x^a  dx` 

∴ `(dI (a))/(da)=[(x^a+1)/(a+1)]1/0` 

∴ `(dI(a))/(da)=1/(a+1)-0` 

∴` (dI(a))/(da)=1/(a+1)`

now , integrate w.r.t a, `I (a)= int 1/(a+1) da` 

`I(a)= log(a+1)+c`            -------- (2) 

where c is constant of integration
put a=0 in eqn (1), 

`I(0)=int_0^1 0  dx=0`

And
From eqn `(2), I (0)=c` 

∴` c=0`

∴` I= log (a+1)`