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Q8 Integrate the rational functions $\frac{x }{( x+1)^2 ( x+2)}$

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

Partial function of this function:

$\frac{x }{( x+1)^2 ( x+2)} = \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+2)}$

$x = A(x-1)(x+2)+B(x+2)+C(x-1)^2$

Now, putting $x=1$ in the above equation, we get

$B =\frac{1}{3}$

By equating the coefficients of $x^2$ and constant term, we get

$A+C=0$

$-2A+2B+C = 0$

then after solving, we get

$A= \frac{2}{9}\ and\ C=\frac{-2}{9}$

Therefore,

$\frac{x}{(x-1)^2(x+2)} = \frac{2}{9(x-1)}+\frac{1}{3(x-1)^2}-\frac{2}{9(x+2)}$

$\int \frac{x}{(x-1)^2(x+2)}dx= \frac{2}{9}\int \frac{1}{(x-1)}dx+\frac{1}{3}\int \frac{1}{(x-1)^2}dx-\frac{2}{9}\int \frac{1}{(x+2)}dx$

$= \frac{2}{9}\log|x-1|+\frac{1}{3}\left ( \frac{-1}{x-1} \right )-\frac{2}{9}\log|x+2|+C$

$\frac{2}{9}\log\left | \frac{x-1}{x+2} \right | -\frac{1}{3(x-1)}+C$

Posted by

Divya Prakash Singh

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