Q. 214.3( 3 Votes )

Evaluate the foll

Answer :

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).


Hence,



Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Evaluate thMathematics - Exemplar

Evaluate the follRD Sharma - Volume 1

Evaluate the follRD Sharma - Volume 1

Evaluate the follRD Sharma - Volume 1

Evaluate the follRD Sharma - Volume 1

If <img widMathematics - Exemplar

Evaluate the follRD Sharma - Volume 1

Find :

Mathematics - Board Papers

Evaluate the follRD Sharma - Volume 1

Evaluate the follRD Sharma - Volume 1