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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals5 Marks
Question
Evaluate the integral $\int_{0}^{2} \frac{d x}{x+4-x^{2}}$ using substitution.
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Answer

Given: $\int_{0}^{2} \frac{d x}{x+4-x^{2}}$
$\int_{0}^{2} \frac{d x}{x+4-x^{2}}=\int_{0}^{2} \frac{d x}{-\left(x^{2}-x-4\right)}$
We can write it as, $\int_{0}^{2} \frac{d x}{-\left(x^{2}-x+\frac{1}{4}-\frac{1}{4}-4\right)}$
= $\int_{0}^{2} \frac{d x}{-\left[\left(x-\frac{1}{2}\right)^{2}-\frac{17}{4}\right]}$
= $\int_{0}^{2} \frac{d x}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}\right]}$
Let $x-\frac{1}{2}=t \Rightarrow d x=d t$
When $x=0, t=-\frac{1}{2}$ and when $\mathrm{x}=2, \mathrm{t}=\frac{3}{2}$
$\Rightarrow \int_{0}^{2} \frac{d x}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}\right]}=\int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{d t}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-(t)^{2}\right]}$
because, $\int \frac{d x}{\left[(a)^{2}-(x)^{2}\right]}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C$
$\Rightarrow \int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{d t}{\left[\left(\frac{\sqrt{17}}{2}\right)^{2}-(t)^{2}\right]}$ = $\left[\frac{1}{2\left(\frac{\sqrt{17}}{2}\right)} \log \frac{\left(\frac{\sqrt{17}}{2}+t\right)}{\frac{\sqrt{17}}{2}-t}\right]_{-\frac{1}{2}}^{\frac{3}{2}}$
= $\frac{1}{\sqrt{17}}\left[\log \frac{\left(\frac{\sqrt{17}}{2}+\frac{3}{2}\right)}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-\log \frac{\left(\frac{\sqrt{17}}{2}-\frac{1}{2}\right)}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right]$
= $\frac{1}{\sqrt{17}}\left[\log \frac{(\sqrt{17}+3)}{\sqrt{17}-3}-\log \frac{(\sqrt{17}-1)}{\sqrt{17}+1}\right]$
= $\frac{1}{\sqrt{17}}\left[\log \left\{\frac{(\sqrt{17}+3)}{\sqrt{17}-3} \times \frac{(\sqrt{17}+1)}{\sqrt{17}-1}\right\}\right]$
= $\frac{1}{\sqrt{17}}\left[\log \left\{\frac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)}\right\}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{17+3+4 \sqrt{17}}{17+3-4 \sqrt{17}}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{20+4 \sqrt{17}}{20-4 \sqrt{17}}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{5+\sqrt{17}}{5-\sqrt{17}}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{(5+\sqrt{17})(5+\sqrt{17})}{(5-\sqrt{17})(5+\sqrt{17})}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{(25+17+10 \sqrt{17})}{25-17}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{(42+10 \sqrt{17})}{8}\right]$
= $\frac{1}{\sqrt{17}} \log \left[\frac{(21+5 \sqrt{17})}{4}\right]$

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