Question

\(\int\limits\sqrt{\dfrac{2+x}{2-x}}^{ }_{ }dx\)

 

Answer 1

\(\int\sqrt{\dfrac{2+x}{2-x}}dx=\int\dfrac{2+x}{\sqrt{4-x^2}}dx\)

Đặt \(x=2sinu\Rightarrow dx=2cosu.du\)

\(\Rightarrow I=\int\dfrac{2+2sinu}{\sqrt{4-4sin^2u}}.2cosu.du=\int\left(2+2sinu\right)du=2u-2cosu+C\)

Trả biến: \(sinu=\dfrac{x}{2}\Rightarrow u=arcsin\left(\dfrac{x}{2}\right)\)

\(x=2sinu\Rightarrow x^2=4sin^2u=4-4cos^2u\Rightarrow cos^2u=1-\dfrac{x^2}{4}\)

\(\Rightarrow cosu=\sqrt{1-\dfrac{x^2}{4}}\)

Vậy \(I=2arcsin\left(\dfrac{x}{2}\right)-\sqrt{4-x^2}+C\)