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Question

Integrate the following functions w.r.t. $$ \mathrm{x} $$.
$$ \int \dfrac{x^{2}-8 x+4}{x^{3}-4 x} d x $$


Solution

$$\begin{array}{l}\int \dfrac{x^{2}-8 x+4}{x^{3}-4 x} d x \\\text{Let }\quad \dfrac{x^{2}-8 x+4}{x^{3}-4 x}=\dfrac{x^{2}-8 x+4}{x\left(x^{2}-4\right)}\end{array}$$
$$=\dfrac{x^{2}-8 x+4}{x(x-2)(x+2)}$$
Let $$ \dfrac{x^{2}-8 x+4}{x(x-2)(x+2)}=\dfrac{A}{x}+\dfrac{B}{x-2}+\dfrac{C}{x+2} $$
$$\begin{array}{r}\Rightarrow \quad x^{2}-8 x+4=A\left(x^{2}-4\right)+\mathrm{B}\left(x^{2}+2 x\right) \\+C\left(x^{2}-2 x\right) \\\Rightarrow \quad x^{2}-8 x+4=(A+B+C) x^{2} \\+(2 B-2 \mathrm{c}) x-4 A\end{array}$$
On comparing,
$$A+B+C=1,2 B-2 C=-8,-4 A=4$$
On solving, $$ A=-1, B=-1, C=3 $$
$$ \therefore \int \dfrac{x^{2}+8 x+4}{x^{3}-4 x} d x=-1 \int \dfrac{1}{x} d x-\int \dfrac{1}{(x-2)} d x $$
$$ =-\log |x|-\log |x-2|+3 \log |x+2|+K $$

Mathematics

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