# `I_n = int_0^1 nx^n /(1+x^n) dx` . Demonstrate` I_n = ln 2-int_0^1 ln (1+x^n) dx` .

sciencesolve | Certified Educator

You may write the given definite integral such that:

`I_n = int_0^1 x*(nx^(n-1))/(1 + x^n)dx`

You may notice that if you come up with the substitution `1 + x^n = y` yields:

`1 + x^n = y => n*x^(n-1) dx = dy`

You may also notice that differentiating the function `ln(1 + x^n)` yields:

`(ln(1 + x^n))' = (nx^(n-1))/(1 + x^n)`

Replacing `(ln(1 + x^n))'` for `(nx^(n-1))/(1 + x^n)` yields:

`int_0^1 x*(ln(1 + x^n))'dx`

Using integration by parts yields:

`f(x) = x => f'(x) = 1`

`g'(x) = (ln(1 + x^n))' => g(x) = ln(1 + x^n)`

`int_0^1 x*(ln(1 + x^n))'dx = x*ln(1 + x^n)|_0^1 - int_0^1 ln(1 + x^n) dx`

`int_0^1 x*(ln(1 + x^n))'dx = 1*ln(1+1) - 0*ln(1+0) - int_0^1 ln(1 + x^n) dx`

`int_0^1 x*(ln(1 + x^n))'dx = ln 2 - int_0^1 ln(1 + x^n) dx`

Hence, testing if `I_n = int_0^1 (nx^(n-1))/(1 + x^n)dx =ln 2 - int_0^1 ln(1 + x^n) dx` , using integration by parts, yields `I_n = ln 2 - int_0^1 ln(1 + x^n) dx` .

oldnick | Student

`I_n= int_0^1 (nx^n)/(1+x^n) dx` `=int_0^1 x xx (n x^(n-1))/(1+x^n) dx` `=int_0^1 x d log(1+x^n)`

Integrating by parts:

`u= x`    `dv= d log(1+x^n)`

`int_0^1 (nx^n)/(1+x^n) dx=` `[x log(1+x^n)]_0^1 - int_0^1 log(1+x^n) dx=`

`=log2 -int_0^1 log(1+x^n)dx`