integration 1/x+4-x² limits 2 to 0
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x+4-x² limits 2 to 0
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Answer:
Step-by-step explanation:
We have to integrate,
[tex]\displaystyle\int^{2}_{0}\frac{1}{x+4-{x}^{2}}\,dx[/tex]
[tex]\displaystyle=\int^{2}_{0}\frac{1}{\displaystyle4+\frac{1}{4}-\frac{1}{4}+2\cdot\frac{1}{2}\cdot\,x-{x}^{2}}\,dx\\[/tex]
[tex]\displaystyle=\int^{2}_{0}\frac{1}{\displaystyle\frac{17}{4}-\left(x-\frac{1}{2}\right)^{2}}\,dx\\[/tex]
[tex]\displaystyle=\int^{2}_{0}\frac{1}{\displaystyle\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}\,dx\\[/tex]
We know that,
[tex]\boxed{\bf{\displaystyle\int\frac{dx}{{a}^{2}-{x}^{2}}=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C}}[/tex]
So,
[tex]\displaystyle=\left[\ln\left|\frac{\displaystyle\frac{\sqrt{17}}{2}+\left(x-\frac{1}{2}\right)}{\displaystyle\frac{\sqrt{17}}{2}-\left(x-\frac{1}{2}\right)}\right|\right]^{2}_{0}\\[/tex]
[tex]\displaystyle=\left[\ln\left|\frac{\displaystyle\sqrt{17}+2x-1}{\displaystyle\sqrt{17}-2x+1}\right|\right]^{2}_{0}\\[/tex]
[tex]\displaystyle=\ln\left|\frac{\displaystyle\sqrt{17}+4-1}{\displaystyle\sqrt{17}-4+1}\right|-\ln\left|\frac{\displaystyle\sqrt{17}+0-1}{\displaystyle\sqrt{17}-0+1}\right|\\[/tex]
[tex]\displaystyle=\ln\left|\frac{\displaystyle\sqrt{17}+3}{\displaystyle\sqrt{17}-3}\right|+\ln\left|\frac{\displaystyle\sqrt{17}+1}{\displaystyle\sqrt{17}-1}\right|\\[/tex]
[tex]\displaystyle=\ln\left|\frac{\displaystyle\sqrt{17}+3}{\displaystyle\sqrt{17}-3}\cdot\frac{\displaystyle\sqrt{17}+1}{\displaystyle\sqrt{17}-1}\right|\\[/tex]
[tex]\displaystyle=\ln\left|\frac{\left(\sqrt{17}+3\right)\left(\sqrt{17}+1\right)}{\left(\sqrt{17}-3\right)\left(\sqrt{17}-1\right)}\right|\\[/tex]
[tex]\displaystyle=\ln\left|\frac{17+4\sqrt{17}+4}{17-4\sqrt{17}+4}\right|\\[/tex]
[tex]\displaystyle=\ln\left|\frac{21+4\sqrt{17}}{21-4\sqrt{17}}\right|\\[/tex]