# Evaluate the foll

Denominator is factorized, so let separate the fraction through partial fraction, hence let    x2 + 1 = A(x – 1)(x + 1) + B(2x + 1)(x + 1) + C(2x + 1)(x – 1)……(ii)

We need to solve for A, B and C. One way to do this is to pick values for x which will cancel each variable.

Put x = 1 in the above equation, we get

12 + 1 = A(1 – 1)(1 + 1) + B(2(1) + 1)(1 + 1) + C(2(1) + 1)(1 – 1)

2 = 0 + 6B + 0 Now put in equation (ii), we get   Now put x = – 1 in equation (ii), we get

( – 1)2 + 1 = A( – 1 – 1)( – 1 + 1) + B(2( – 1) + 1)( – 1 + 1) + C(2( – 1) + 1)( – 1 – 1)

2 = 0 + 0 + 2C

C = 1

We put the values of A, B, and C values back into our partial fractions in equation (i) and replace this as the integrand. We get   Split up the integral, Let substitute

u = x – 1 du = dx,

y = x + 1 dy = dx and

z = 2x + 1 dz = 2dx so the above equation becomes, On integrating we get Substituting back, we get Note: the absolute value signs account for the domain of the natural log function (x>0).

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