# Evaluate the foll

Denominator is factorised, so let separate the fraction through partial fraction, hence let     We need to solve for A, B, C and D. We will equate the like terms we get,

C = 0………….(iii)

A + D = 1 A = 1 – D………(iv)

2A + B + C = 1

2(1 – D) + B + 0 = 1 (from equation (iii) and (iv))

B = 2D – 1……….(v)

2B + D = 1

2(2D – 1) + D = 1 (from equation (v), we get

4D – 2 + D = 1

5D = 3 ……………(vi)

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

u = x2 + 1 du = 2xdx,

y = x + 2 dy = dx, so the above equation becomes, On integrating we get (the standard integral of )

Substituting back, we get Note: the absolute value signs account for the domain of the natural log function (x>0).

Hence, Rate this question :

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