The integral 24π∫02(2−x2)dx(2+x2)4+x4\dfrac{24}\pi\displaystyle\int_0^{\sqrt2}\frac{\left(2-x^2\right)dx}{\left(2+x^2\right)\sqrt{4+x^4}}π24∫02(2+x2)4+x4(2−x2)dx is equal to ______.
Official answer: 3
Divide by x2x^2x2. (2−x2) dx(2+x2)4+x4=−duuu2−4\dfrac{(2-x^2)\,dx}{(2+x^2)\sqrt{4+x^4}}=\dfrac{-du}{u\sqrt{u^2-4}}(2+x2)4+x4(2−x2)dx=uu2−4−du with u=x+2xu=x+\dfrac2xu=x+x2.
Limits. x→0x\to0x→0: u→∞u\to\inftyu→∞; x=2x=\sqrt2x=2: u=22u=2\sqrt2u=22. So the integral is ∫22∞duuu2−4=12[sec−1u2]22∞=12(π2−π4)=π8\displaystyle\int_{2\sqrt2}^\infty\frac{du}{u\sqrt{u^2-4}}=\frac12\left[\sec^{-1}\frac u2\right]_{2\sqrt2}^\infty=\frac12\left(\frac\pi2-\frac\pi4\right)=\frac\pi8∫22∞uu2−4du=21[sec−12u]22∞=21(2π−4π)=8π.
Value. 24π⋅π8=3\dfrac{24}\pi\cdot\dfrac\pi8=3π24⋅8π=3.
Answer: 3