Question

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$\int\limits {\frac{dx}{\sqrt{8+2x-x^2} } } \,$

Answer 1

$\large\boxed{\begin{array}{l}\rm 8+2x-x^2=-(x^2-2x-8)\\\rm-(x^2-2x+1-9)=-(x^2-2x+1)+9\\\rm 8+2x-x^2=9-(x-1)^2\\\underline{\sf Fac_{\!\!\!,}\!~\!a}\\\rm u=x-1\longrightarrow du=dx\\\rm 9-(x-1)^2=9-u^2\\\displaystyle\rm\int\dfrac{dx}{8+2x-x^2}=\int\dfrac{du}{\sqrt{9-u^2}}\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\int\!\!\dfrac{du}{\sqrt{9-u^2}}=\int\dfrac{du}{\sqrt{9\cdot\bigg(1-\bigg[\dfrac{u}{3}\bigg]^2\bigg)}}=\dfrac{1}{3}\int\dfrac{du}{\sqrt{1-\bigg[\dfrac{u}{3}\bigg]^2}}\\\underline{\sf fac_{\!\!\!,}\!~\!a}\\\rm t=\dfrac{u}{3}\longrightarrow dt=\dfrac{1}{3}\,du\\\\\displaystyle\rm\int\dfrac{du}{\sqrt{9-u^2}}=\dfrac{1}{3}\cdot\dfrac{1}{3}\int\dfrac{dt}{\sqrt{1-t^2}}=\dfrac{1}{9}\,arc sen(t)+k \end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\rm\int\dfrac{du}{\sqrt{9-u^2}}=\dfrac{1}{9}\,arcsen\bigg(\dfrac{u}{3}\bigg)+k\\\\\displaystyle\rm\int\dfrac{dx}{\sqrt{8+2x-x^2}}=\dfrac{1}{9}\,arcsen\bigg(\dfrac{x-1}{3}\bigg)+k\end{array}}$