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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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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