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Q13  Integrate the rational functions \frac{2}{(1-x)(1+ x^2)}

Answers (1)


Given function \frac{2}{(1-x)(1+ x^2)}

can be rewritten as \frac{2}{(1-x)(1+ x^2)} = \frac{A}{(1-x)}+\frac{Bx+C}{1+x^2}

2 =A(1+x^2)+(Bx+C)(1-x)                           ....................(1)

2 =A +Ax^2 +Bx-Bx^2+C-Cx

Now, equating the coefficient of x^2, x, and constant term, we get

A-B= 0B-C = 0, and A+C =2

Solving these equations, we get

A=1, B=1,\ and\ C=1


\therefore \frac{2}{(1-x)(1+ x^2)} = \frac{1}{(1-x)}+\frac{x+1}{1+x^2}

\implies \int \frac{2}{(1-x)(1+ x^2)}dx =\int \frac{1}{(1-x)} dx+ \int \frac{x}{1+x^2}dx +\int \frac{1}{1+x^2}dx= -\int \frac{1}{x-1}dx +\frac{1}{2}\int \frac{2x}{1+x^2}dx +\int\frac{1}{1+x^2}dx

=-\log|x-1| +\frac{1}{2}\log|1+x^2| +\tan^{-1}x+C

Posted by

Divya Prakash Singh

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