Chapter-wise Solutions

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This blog contains solutions to unsolved problems on the book. Some questions have complete solutions and explanations and some easier questions have hints to solve the problem. I assume that you've gone through the worked out problems and theory given in the book first. Feel free to leave a comment if you have any doubt or if you found a problem not done right.
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Saturday 20 April 2019

Answer to question 5.1.14

Answer to 5.1.14

We have the question,

Since the fraction is improper, we need to single out the integral part(Refer 5.1.3 solved example in the textbook). Dividing numerator by denominator, we obtain :


Hence using this, our integral becomes :


Expand the remaining proper fraction into simple ones using partial fraction :


Hence :

Putting x = -1, we get 2A = -1 + 1 +2, or, A = 1

Now, expanding and comparing coefficients of similar terms on both sides and solving the equation, we get B = -2 and C = 1.

So,we have :

So, the integral becomes :


Integrating, we get the answer as :