Mathematics

Integrate the following function defined over
$$\frac{x^2+4x-1}{x^3-x}$$


SOLUTION
Let $$\dfrac{x^2+4x-1}{x^3-x}=\dfrac{x^2+4x-1}{x(x^2-1)}=\dfrac{A}{x}+\dfrac{B}{x+1}+\dfrac{C}{x-1}$$
$$\therefore \dfrac{x^2+4x-1}{x(x^2-1)}=\dfrac{A(x^2-1)+B(x^2-x)+c(x^2+x)}{x(x^2-1)}$$
So, $$x^2+4x-1=A(x^2-1)+B(x^2-x)+C(x^2+x)$$
$$\therefore A+B+C=1$$ …..(i)
$$C-B=4$$ ……….(ii)
$$-A=-1$$ …….(iii)
equation (i) $$+$$ equation (ii) $$+$$ equation (iii), gives $$2C=4$$
$$\Rightarrow C=2$$
$$\therefore A=1, C=2, B=-2$$
$$\therefore \dfrac{x^2+4x-1}{x^3-x}=\dfrac{1}{x}-\dfrac{2}{x+1}+\dfrac{2}{x-1}$$
$$\therefore \displaystyle\int \dfrac{x^2+4x-1}{x^3-x}dx=\displaystyle\int \dfrac{1}{x}dx-\displaystyle\int \dfrac{2}{x+1}dx+\displaystyle\int \dfrac{2}{x-1}dx$$
$$=log|x|-2log(x+1)|+2log|(x-1)|$$
$$\displaystyle\int \dfrac{x^2+4x-1}{x^2-x}dx=log\left|\left\{\dfrac{x(x-1)^2}{(x+1)^2}\right\}\right|$$
$$\therefore \displaystyle\int \dfrac{x^2+4x-1}{x^3-x}dx=log\left|\dfrac{x(x-1)^2}{(x+1)^2}\right|$$.
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