Q. 144.6( 5 Votes )

Evaluate the foll

Answer :

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,



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