Evaluate the following intregals: 2x^2+7x-3 x^2(2x+1) dx

Question

Evaluate the following intregals:
$\int\frac{2\text{x}^2+7\text{x}-3}{\text{x}^2(2\text{x}+1)}\text{ dx}$

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Answer

we have, $\text{I}=\int\frac{2\text{x}^2+7\text{x}-3}{\text{x}^2(2\text{x}+1)}\text{ dx}$ $\Rightarrow\int\frac{2\text{x}^2+7\text{x}-3}{\text{x}^2(2\text{x}+1)}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}^2}+\frac{\text{C}}{2\text{x}+1}$ $\Rightarrow\int\frac{2\text{x}^2+7\text{x}-3}{\text{x}^2(2\text{x}+1)}=\frac{\text{A}(\text{x})(2\text{x}+1)+\text{B}(2\text{x}+1)+\text{Cx}^2}{\text{x}^2(2\text{x}+1)}$ $\Rightarrow2\text{x}^2+7\text{x}-3=\text{A}(2\text{x}^2+\text{x})+\text{B}(2\text{x}+1)+\text{Cx}^2$ $\Rightarrow2\text{x}^2+7\text{x}-3=(2\text{A}+\text{C})\text{x}^2+(\text{A}+2\text{})\text{x}+\text{B}$ Equating coefficient of like terms 2A + C = 2 ...(1) A + 2B = 7 ...(2) B = -3 ...(3) Solving (1), (2) and (3), we get A = 13 B = -3 C = -24$\therefore\frac{2\text{x}^2+7\text{x}-3}{\text{x}^2(2\text{x}+1)}=\frac{13}{\text{x}}-\frac{3}{\text{x}^2}-\frac{24}{2\text{x}+1}$
$\Rightarrow\text{I}=13\int\frac{\text{dx}}{\text{x}}-3\int\text{x}^{-2}\text{dx}-24\int\frac{\text{dx}}{2\text{x}+1}$
$=13\log|\text{x}|+\frac{3}{\text{x}}-24\frac{\log|2\text{x}+1|}{2}+\text{C}$
$=13\log|\text{x}|+\frac{3}{\text{x}}-12\log|2\text{x}-1|+\text{C}$

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