Evaluate the following integrals: 1 2x^2-x-1 dx - Vidyadip

Question

Evaluate the following integrals:$\int\frac{1}{2\text{x}^2-\text{x}-1}\text{dx}$

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Answer

Let $\text{I}=\int\frac{1}{2\text{x}^2-\text{x}-1}\text{dx}$
$=\frac{1}{2}\int\frac{1}{\text{x}^2-\frac{\text{x}}{2}-\frac{1}{2}}\text{dx}$
$=\frac{1}{2}\int\frac{1}{\text{x}^2-2\text{x}\times\frac{1}{4}+\big(\frac{1}{4}\big)^2-\big(\frac{1}{4}\big)^2-\frac{1}{2}}\text{dx}$
$=\frac{1}{2}\int\frac{1}{\big(\text{x}-\frac{1}{4}\big)^2-\frac{9}{16}}\text{dx}$
Let $\text{x}-\frac{1}{4}=\text{t}$
$\Rightarrow\text{dx = dt}$
$\text{I}=\frac{1}{2}\int\frac{1}{\text{t}^2-\big(\frac{3}{4}\big)^2}\text{dt}$
$\text{I}=\frac{1}{2}\times\frac{1}{2\times\big(\frac{3}{4}\big)}\log\Bigg|\frac{\text{t}-\frac{3}{4}}{\text{t}+\frac{3}{4}}\Bigg|+\text{C} $ $\Big[\text{Since,}\int\frac{1}{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\log\bigg|\frac{\text{x}-\text{a}}{\text{x+a}}\bigg|+\text{C}\Big]$
$\text{I}=\frac{1}{3}\log\Bigg|\frac{\text{x}-\frac{1}{4}-\frac{3}{4}}{\text{x}-\frac{1}{4}+\frac{3}{4}}\Bigg|+\text{C}$
$\text{I}=\frac{1}{3}\log\bigg|\frac{\text{x}-1}{2\text{x}+1}\bigg|+\text{C}$

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