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Question:

$\int_{0}^{2} \frac{d x}{x+4-x^{2}}$

Solution:

$\int_{0}^{2} \frac{d x}{x+4-x^{2}}=\int_{0}^{2} \frac{d x}{-\left(x^{2}-x-4\right)}$

$=\int_{0}^{2} \frac{d x}{-\left(x^{2}-x+\frac{1}{4}-\frac{1}{4}-4\right)}$

$=\int_{0}^{2} \frac{d x}{-\left[\left(x-\frac{1}{2}\right)^{2}-\frac{17}{4}\right]}$

$=\int_{0}^{2} \frac{d x}{\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}$

Let $x-\frac{1}{2}=t \Rightarrow d x=d t$

When $x=0, t=-\frac{1}{2}$ and when $x=2, t=\frac{3}{2}$

$\therefore \int_{0}^{2} \frac{d x}{\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}=\int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{d t}{\left(\frac{\sqrt{17}}{2}\right)^{2}-t^{2}}$

$=\left[\frac{1}{2\left(\frac{\sqrt{17}}{2}\right)} \log \frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\right]_{\frac{1}{2}}^{\frac{3}{2}}$

$=\frac{1}{\sqrt{17}}\left[\log \frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-\frac{\log \frac{\sqrt{17}}{2}-\frac{1}{2}}{\log \frac{\sqrt{17}}{2}+\frac{1}{2}}\right]$

$=\frac{1}{\sqrt{17}}\left[\log \frac{\sqrt{17}+3}{\sqrt{17}-3}-\log \frac{\sqrt{17}-1}{\sqrt{17}+1}\right]$

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

$=\frac{1}{\sqrt{17}} \log \left[\frac{17+3+4 \sqrt{17}}{17+3-4 \sqrt{17}}\right]$

$=\frac{1}{\sqrt{17}} \log \left[\frac{20+4 \sqrt{17}}{20-4 \sqrt{17}}\right]$

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

$=\frac{1}{\sqrt{17}} \log \left[\frac{(5+\sqrt{17})(5+\sqrt{17})}{25-17}\right]$

$=\frac{1}{\sqrt{17}} \log \left[\frac{25+17+10 \sqrt{17}}{8}\right]$

$=\frac{1}{\sqrt{17}} \log \left(\frac{42+10 \sqrt{17}}{8}\right)$

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

 

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