Question

Evaluate the following definite integral:

${\int }_0^2\frac{1}{4+x-x^2}dx$

Answer 1

Consider, $I={\int }_0^2\frac{1}{4+x-x^2}dx$

$=-{\int }_0^2\frac{1}{x^2-x-4}dx$

$=-{\int }_0^2\frac{1}{x^2-x-\frac{1}{4}-\frac{17}{4}}dx$

$=-{\int }_0^2\frac{1}{{(x-\frac{1}{2})}^2-{\left(\frac{\sqrt{17}}{2}\right)}^2}dx$

$\Rightarrow I={\int }_0^2\frac{1}{{\left(\frac{\sqrt{17}}{2}\right)}^2-{(x-\frac{1}{2})}^2}dx$

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

$\Rightarrow I=\frac{1}{\sqrt{17}}\{\frac{\sqrt{17}+3}{\sqrt{17}-3}-\log \frac{\sqrt{17}-1}{\sqrt{17}+1}\}$

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

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

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