Mathematics

Evaluate: $$\displaystyle\int {{\dfrac{1}  {{x^2} - x - 1}}} dx$$


SOLUTION
$$=\displaystyle\int \dfrac{1}{x^2-x-1}dx$$

$$=\displaystyle\int \dfrac{1}{x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-1}dx$$

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

$$=\displaystyle\int \dfrac{1}{\left(x-\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2}dx$$

Let $$x-\dfrac{1}{2}=u$$

Here $$a=\dfrac{\sqrt{5}}{2}$$

$$dx=du$$

$$=\displaystyle\int \dfrac{1}{u^2-a^2}du$$ [by formula $$\displaystyle\int \dfrac{dx}{x^2-a^2}=\dfrac{1}{2a}log\left|\dfrac{x-a}{x+a}\right|+c$$]

$$=\dfrac{1}{2a}log\left|\dfrac{u-a}{u+a}\right|+c$$ (putting a and u)

$$=\dfrac{1}{2\times \dfrac{\sqrt{5}}{2}}log\left|\dfrac{x-\dfrac{1}{2}-\dfrac{\sqrt{5}}{2}}{x-\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}}\right|+c$$

$$=\dfrac{1}{\sqrt{5}} log\left|\dfrac{2x-1-\sqrt{5}}{2x-1+\sqrt{5}}\right|+c$$.
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Subjective Medium Published on 17th 09, 2020
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