Mathematics

Evaluate: $$\displaystyle \int_{0}^{2}\dfrac {1}{4+x-x^2}dx$$


SOLUTION
$$I=\displaystyle \int_0^2 \dfrac {1}{4+x-x^2}dx$$ 

$$I=-\displaystyle \int_0^2 \dfrac {1}{x^2 -x-4}dx $$ 

$$I=-\displaystyle \int_0^2\dfrac {1}{\left (x-\dfrac {1}{2}\right)^2-\left (\dfrac {\sqrt {17}}{2}\right)^2}dx$$

$$\ I=\displaystyle \int_0^2 \dfrac {1}{\left (\dfrac {\sqrt {17}}{2}\right)^2-\left (x-\dfrac {1}{2}\right)^2}dx $$ 

$$I=\dfrac {1}{\sqrt {17}} \left [\log \left (\dfrac {\sqrt {17}+2x-1}{\sqrt {17} -2x+1}\right) \right]_0^2$$

$$\ I=\dfrac {1}{\sqrt {17}}\left\{\log \dfrac {\sqrt {17}+3}{\sqrt {17}-3} -\log \dfrac {\sqrt {17}-1}{\sqrt {17}+1} \right\}$$ 

$$I=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {20+4\sqrt {17}}{20-4\sqrt {17}} \times \dfrac {20+4\sqrt {17}}{20+4\sqrt {17}}\right)$$

$$I=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {672+160\sqrt {17}}{128}\right)$$ 

$$I=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {21+5\sqrt {17}}{4}\right)$$
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Subjective Medium Published on 17th 09, 2020
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