Question

Evaluate the integrals using substitution.
${\int }_0^2\frac{1}{x+4-x^2}dx.$

Answer 1

${\int }_0^2\frac{1}{x+4-x^2}dx={\int }_0^2\frac{1}{x+4-x^2-(\frac{1}{2}{)}^2+(\frac{1}{2}{)}^2}dx={\int }_0^2\frac{1}{4+\frac{1}{4}-\left[\right(x^2-x+\frac{1}{4}\left)\right]}dx={\int }_0^2\frac{1}{4+\frac{1}{4}-(x-\frac{1}{2}{)}^2}dx={\int }_0^2\frac{1}{(\frac{\sqrt{17}}{2}{)}^2-(x-\frac{1}{2}{)}^2}dx$

$=\frac{1}{2.\frac{\sqrt{17}}{2}}{\left[log\left|\frac{\frac{\sqrt{17}}{2}+x-\frac{1}{2}}{\frac{\sqrt{17}}{2}-x+\frac{1}{2}}\right|\right]}_0^2(\because \int \frac{dx}{a^2-x^2}=\frac{1}{2a}log\left|\frac{a+x}{a-x}\right|)=\frac{1}{\sqrt{17}}{\left[log\left|\frac{\sqrt{17}+2x-1}{\sqrt{17}-2x+1}\right|\right]}_0^2=\frac{1}{\sqrt{17}}\left[log|\frac{\sqrt{17}+3}{\sqrt{17}-3}-log\left|\frac{\sqrt{17}-1}{\sqrt{17}+1}\right||\right]=\frac{1}{\sqrt{17}}log|\frac{\sqrt{17}+3}{\sqrt{17}-3}+\frac{\sqrt{17}-1}{\sqrt{17}+1}|(\because logm-logn=log\frac{m}{n})$

$=\frac{1}{\sqrt{17}}log\left|\frac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)}\right|=\frac{1}{\sqrt{17}}log\left|\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\right|=\frac{1}{\sqrt{17}}log|\frac{5+\sqrt{17}}{5-\sqrt{17}}\times \frac{5+\sqrt{17}}{5+\sqrt{17}}|$

To remove square root term from denominator, we multiply by its conjugate i.e.,$5+\sqrt{17}$ in both numerator and denominator.

$=\frac{1}{\sqrt{17}}log\left|\frac{(5+\sqrt{17}{)}^2}{(5{)}^2-(\sqrt{17}{)}^2}\right|[\because (a-b\left)\right(a+b)=a^2-b^2]=\frac{1}{\sqrt{17}}log\left|\frac{25+17+10\sqrt{17}}{25-17}\right|\frac{1}{\sqrt{17}}log\left|\frac{42+10\sqrt{17}}{8}\right|=\frac{1}{\sqrt{17}}log\left|\frac{21+5\sqrt{17}}{4}\right|$