Mathematics

$$\int {\dfrac{{5{x}}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}} \,\,dx$$


SOLUTION
$$\int\dfrac{5x}{(x+1)(x^2-4)}$$

$$\dfrac{5x}{(x+1)(x^2-4)}=\dfrac{5x}{(x+1)(x-2)(x+2)}$$

$$\dfrac{5x}{(x+1)(x^2-4)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x+2)}$$

$$\dfrac{5x}{(x+1)(x^2-4)}=\dfrac{A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)}{(x+1)(x-2)(x+2)}$$

By cancelling denominator
$$5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)$$
Putting $$x=-1$$, in ( 1 )
$$5\times (-1)=A(-1-2)(-1+2)+B(-1+1)(-1+2)+C(-1+1)(-1-2)$$
$$-5=A(-3)(1)$$
$$\Rightarrow$$  $$A=\dfrac{5}{3}$$
Similarly, putting $$x=2,$$ in ( 1 )
$$5(2)=A(2-2)(2+2)+B(2+1)(2+2)+c(2+1)(2-2)$$
$$10=A\times 0+B(3)(4)+C\times 0$$
$$10=12B$$
$$\Rightarrow$$  $$B=\dfrac{10}{12}=\dfrac{5}{6}$$
Similarly putting $$x=-2,$$ in ( 1 )
$$5(-2)=A(-2-2)(-2+2)+B(-2+1)(-2+2)+C(-2+1)(-2-2)$$
$$-10=A\times 0+\times 0+C(-1)(-4)$$
$$-10=4C$$
$$C=\dfrac{-10}{4}=\dfrac{-5}{2}$$
Therefore
$$\int\dfrac{5x}{(x+1)(x^2-4)}=\int\left(\dfrac{A}{x+1}+\dfrac{B}{x-2}+\dfrac{C}{x+2}\right)dx$$

                                  $$=\dfrac{5}{3}\int\dfrac{dx}{x+1}dx+\dfrac{5}{6}\int\dfrac{dx}{x-2}dx-\dfrac{5}{2}\int\dfrac{dx}{x+2}$$

                                  $$=\dfrac{5}{3}log|x+1|-\dfrac{5}{2}\log|x+2|+\log|x-2|+c$$

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