Mathematics

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


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

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

$$=-\displaystyle \int_0^2\dfrac {1}{x^2-x-\dfrac {1}{4}-\dfrac {{17}}{4}}dx$$

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


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


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


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


$$=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {52+12\sqrt {17}}{18-2\sqrt {17}} \times \dfrac {18+2\sqrt {17}}{18+2\sqrt {17}}\right)$$


$$\Rightarrow \ I=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {1344+320\sqrt {17}}{256}\right)$$


$$=\dfrac {1}{\sqrt {17}}\log \left (\dfrac {21+5\sqrt {17}}{4}\right)$$

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