Q. 195.0( 1 Vote )

Evaluate the following integral:


Answer :

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




5x2 – 1 = A(x – 1)(x + 1) + Bx(x + 1) + Cx(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 = 0 in the above equation, we get


5(0)2 – 1 = A(0 – 1)(0 + 1) + B(0)(0 + 1) + C(0)(0 – 1)


A = 1


Now put x = 1 in equation (ii), we get


5(1)2 – 1 = A(1 – 1)(1 + 1) + B(1)(1 + 1) + C(1)(1 – 1)


4 = 0 + 2B + 0


B = 2


Now put x = – 1 in equation (ii), we get


5( – 1)2 – 1 = A( – 1 – 1)( – 1 + 1) + B( – 1)( – 1 + 1) + C( – 1)( – 1 – 1)


4 = 0 + 0 + 2C


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 – 1 du = dx,


y = x + 1 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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