Evaluate the following intregals: 5x (x+1)(x^2-4) dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}\text{dx}$

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Answer

$\int\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}=\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}$
Let $\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}=\frac{\text{A}}{(\text{x}+1)}+\frac{\text{B}}{(\text{x}+2)}+\frac{\text{C}}{(\text{x}-2)}$
$5\text{x}=\text{A}(\text{x}+2)(\text{x}-2)+\text{B}(\text{x}+1)(\text{x}-2)\\+\text{C}(\text{x}+1)(\text{x}+2)\ \dots(1)$
Substituting x = -1, -2 and 2 respectively in equation (1), we obtain
$\text{A}=\frac{5}{3},\text{B}=\frac{5}{2},\text{and }\text{C}=\frac{5}{6}$
$\therefore\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}=\frac{5}{3(\text{x}+1)}-\frac{5}{2(\text{x}+2)}+\frac{5}{6(\text{x}-2)}$
$\Rightarrow\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}\ \text{dx}=\frac{5}{3}\int\frac{1}{(\text{x}+1)}\ \text{dx}-\frac{5}{2}\int\frac{1}{(\text{x}+2)}\\\ \text{dx}+\frac{5}{6}\int\frac{1}{(\text{x}-2)}\ \text{dx}$
$=\frac{5}{3}\log|\text{x}+1|-\frac{5}{2}\log|\text{x}+2|+\frac{5}{6}\log|\text{x}-2|+\text{C}$

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