# Evaluate the foll

Denominator is factorized, so let separate the fraction through partial fraction, hence let

x2 + 6x – 8 = A(x – 2)(x + 2) + Bx(x + 2) + Cx(x – 2)……(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 = 0 in the above equation, we get

02 + 6(0) – 8 = A(0 – 2)(0 + 2) + B(0)(0 + 2) + C(0)(0 – 2)

– 8 = – 4A + 0 + 0

A = 2

Now put x = 2 in equation (ii), we get

22 + 6(2) – 8 = A(2 – 2)(2 + 2) + B(2)(2 + 2) + C(2)(2 – 2)

8 = 0 + 8B + 0

B = 1

Now put x = – 2 in equation (ii), we get

( – 2)2 + 6( – 2) – 8 = A(( – 2) – 2)(( – 2) + 2) + B( – 2)(( – 2) + 2) + C( – 2)(( – 2) – 2)

– 16 = 0 + 0 + 8C

C = – 2

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 – 2 du = dx,

y = x + 2 dy = dx, so the above equation becomes,

On integrating we get

Substituting back, we get

Applying logarithm rule, we get

Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence,

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