Evaluate the following integral:
Multiply numerator and denominator by xn – 1, we get
Let xn = t ⇒ nxn – 1dx = dt
So the above equation becomes,
The denominator is factorized, so let separate the fraction through partial fraction, hence let
⇒ 1 = A(t + 1) + Bt……(ii)
Put t = 0 in above equations we get
1 = A(0 + 1) + B(0)
⇒ A = 1
Now put t = – 1 in equation (ii) we get
1 = A( – 1 + 1) + B( – 1)
⇒ B = – 1
We put the values of A and B values back into our partial fractions in equation (i) and replace this as the integrand. We get
Split up the integral,
Let substitute
u = t + 1 ⇒ du = dt, so the above equation becomes,
On integrating we get
Substituting back the values of u, we get
Substituting back the values of t, we get
Applying the logarithm rules, we get
Note: the absolute value signs account for the domain of the natural log function (x>0).
Hence,