Question

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

Answer 1

${\int }_0^2\frac{dx}{x+4-x^2}={\int }_0^2\frac{dx}{-(x^2-x-4)}$
$={\int }_0^2\frac{dx}{-(x^2-x+\frac{1}{4}-\frac{1}{4}-4)}$
$={\int }_0^2\frac{dx}{-[{(x-\frac{1}{2})}^2-\frac{17}{4}]}$
$={\int }_0^2\frac{dx}{{\left(\frac{\sqrt{17}}{2}\right)}^2-{(x-\frac{1}{2})}^2}$
Let $x-\frac{1}{2}=t\Rightarrow dx=dt$
When $x=0,t=-\frac{1}{2}$ and when $x=2,t=\frac{3}{2}$
$\therefore {\int }_0^2\frac{dx}{{\left(\frac{\sqrt{17}}{2}\right)}^2-{(x-\frac{1}{2})}^2}={\int }_{-\frac{1}{2}}^{\frac{3}{2}}\frac{dt}{{\left(\frac{\sqrt{17}}{2}\right)}^2-t^2}$
$={\left[\frac{1}{2\left(\frac{\sqrt{17}}{2}\right)}\log \frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\right]}_{-\frac{1}{2}}^{\frac{3}{2}}$
$=\frac{1}{\sqrt{17}}[\log \frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-\log \frac{\frac{\sqrt{17}}{2}-\frac{1}{2}}{\frac{\sqrt{17}}{2}+\frac{1}{2}}]$
$=\frac{1}{\sqrt{17}}[\log \frac{\sqrt{17}+3}{\sqrt{17}-3}-\log \frac{\sqrt{17}-1}{\sqrt{17}+1}]$
$=\frac{1}{\sqrt{17}}[\log \frac{\sqrt{17}+3}{\sqrt{17}-3}\times \frac{\sqrt{17}+1}{\sqrt{17}-1}]$
$=\frac{1}{\sqrt{17}}\log \left[\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}}\right]$
$=\frac{1}{\sqrt{17}}\log \left[\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\right]$
$=\frac{1}{\sqrt{17}}\log \left(\frac{5+\sqrt{17}}{5-\sqrt{17}}\right)$
$=\frac{1}{\sqrt{17}}\log \left[\frac{(5+\sqrt{17})(5+\sqrt{17})}{25-17}\right]$
$=\frac{1}{\sqrt{17}}\log \left[\frac{25+17+10\sqrt{17}}{8}\right]$
$=\frac{1}{\sqrt{17}}\log \left(\frac{42+10\sqrt{17}}{8}\right)$
$=\frac{1}{\sqrt{17}}\log \left(\frac{21+5\sqrt{17}}{4}\right)$