Find the value of

take xⁿ +1 as t.

differentiating, n x^n-1 dx = dt

substitute.

we get   dt/n x^n-1 x (t)

=dt/n xⁿ t

=dt/n t (t-1)

now you can integrate using partial fractions.

At + B(t-1)=1

A=-1, B=1

so, we get (-1) dt/t +dt/(t-1)

= -ln t + ln(t-1)

= - ln (xⁿ + 1) + ln xⁿ +c

=ln [(xⁿ + 1)/xⁿ] = ln [1+1/xⁿ] +c

is it  ln(x)  - { [ ln|xⁿ+1| ] / n }                                                                                                                                                                                                                                                                                                                                                            

let xpowern+1=t.

on diffrentiating we getnxpowern/x=dt/dx.

dx/x=dt/nxpowern.

=integral 1/t(t-1).Using partial fractions we get the answer as

log(xpowern)-log(xpowern+1).

Put logx=t.

dx/x = dt

so xⁿ = e^(nt)

the integral becomes.....dt/(e^nt + 1)

Multiply the numerator and denominator by e^-nt so that it gets converted to the form f^'(x)/f(x).

multiply Nr. and Dr. bu x^(n-1) and then put x^n = t