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Q16 Integrate the rational functions \frac{1}{x ( x^n+1)} 

[Hint: multiply numerator and denominator by x ^{n-1} and put x ^n = t ]

Answers (1)


Given function \frac{1}{x ( x^n+1)}

Applying Hint multiplying numerator and denominator by x^{n-1} and putting x^n =t

\frac{1}{x ( x^n+1)} = \frac{x^{n-1}}{x^{n-1}x(x^n+1)} = \frac{x^{n-1}}{x^n(x^n+1)}

Putting x^n =t 

\therefore x^{n-1}dx =dt

can be rewritten as \int \frac{1}{x ( x^n+1)}dx =\int \frac{x^{n-1}}{x^n(x^n+1)}dx = \frac{1}{n} \int \frac{1}{t(t+1)}dt

Partial fraction of above equation,

\frac{1}{t(t+1)} =\frac{A}{t}+\frac{B}{(t+1)}

1 = A(1+t)+Bt                                      ................(1)

Now, substituting t = 0,-1 in equation (1), we get

A=1\ and\ B=-1

\therefore \frac{1}{t(t+1)} = \frac{1}{t}- \frac{1}{(1+t)}

\implies \int \frac{1}{x(x^n+1)}dx = \frac{1}{n} \int \left \{ \frac{1}{t}-\frac{1}{(t+1)} \right \}dx

= \frac{1}{n} \left [ \log|t| -\log|t+1| \right ] +C

= -\frac{1}{n} \left [ \log|x^n| -\log|x^n+1| \right ] +C

= \frac{1}{n} \log|\frac{x^n}{x^n+1}| +C 


Posted by

Divya Prakash Singh

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