Evaluvate the following intregals: x^2+x-1 x^2+x-6 dx - Vidyadip

Question

Evaluvate the following intregals:
$\int\frac{\text{x}^2+\text{x}-1}{\text{x}^2+\text{x}-6}\ \text{dx}$

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Answer

$\int\frac{\text{x}^2+\text{x}-1}{\text{x}^2+\text{x}-6}\ \text{dx}$
$=\int\Big(\frac{\text{x}^2+\text{x}-6+6-1}{\text{x}^2+\text{x}-6}\Big)\text{dx}$
$=\int\Big(\frac{\text{x}^2+\text{x}-6}{\text{x}^2+\text{x}-6}\Big)\ \text{dx}+5\int\frac{1}{\text{x}^2+\text{x}-6}\ \text{dx}$
$=\int\text{dx}+5\int\frac{1}{\text{x}^2+3\text{x}-2\text{x}-6}\text{dx}$
$=\int\text{dx}+5\int\frac{1}{\text{x}(\text{x}+3)-2(\text{x}+3)}\ \text{dx}$
$=\int\text{dx}+5\int\frac{1}{\text{x}(\text{x}+3)-2(\text{x}+3)}\ \text{dx}$
$=\int\text{dx}+5\int\frac{1}{(\text{x}-2)(\text{x}+3)}\ \text{dx}\ \dots(1)$
Let $\frac{1}{(\text{x}-2)(\text{x}+3)}=\frac{\text{A}}{\text{x}-2}+\frac{\text{B}}{\text{x}+3}$
$\Rightarrow\frac{1}{(\text{x}-2)(\text{x}+3)}=\frac{\text{A}(\text{x}+3)+\text{B}(\text{x}-2)}{(\text{x}-2)(\text{x}+3)}$
$\Rightarrow1=\text{A}(\text{x}+3)+\text{B}(\text{x}-2)\ \dots(2)$
putting $\text{x}+3=0\text{ or }\text{x}=-3$ in eq (2)
$\Rightarrow1=\text{A}\times0+\text{B}(-3-2)$
$\Rightarrow\text{B}=-\frac{1}{5}$
Putting $\text{x}-2=0\text{ or }\text{x}=2$ in eq (2)
$\Rightarrow1=\text{A}(2+3)+\text{B}\times0$
$\Rightarrow\text{A}=\frac{1}{5}$
$\therefore\frac{1}{(\text{x}-2)(\text{x}+3)}=\frac{1}{5}(\text{x}-2)-\frac{1}{5}(\text{x}+3)$
$\Rightarrow\int\frac{1}{(\text{x}-2)(\text{x}+3)}\text{ dx}=\frac{1}{5}\int\frac{\text{dx}}{\text{x}-2}-\frac{1}{5}\int\frac{\text{dx}}{\text{x}+3}$
$=\frac{1}{5}\ln|\text{x}-2|-\frac{1}{5}\ln|\text{x}+3|+\text{C}$
$=\frac{1}{5}\ln\Big|\frac{\text{x}-2}{\text{x}+3}\Big|+\text{C}\ ....(3)$
From eq (1) and (3)
$\therefore\int\Big(\frac{\text{x}^2+\text{x}-1}{\text{x}^2+\text{x}-61}\Big)\text{dx}=\text{x}+\frac{5}{5}\ln\Big|\frac{\text{x}-2}{\text{x}+3}\Big|+\text{C}$
$=\text{x}+\ln|\text{x}-2|-\ln|\text{x}+3|+\text{C}$

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