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dydx=y=(x+2)2x2+2\frac{dy}{dx}=y=\left(x+2\right)^2\sqrt{x^2+2}

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Final answer to the exercise

ln⁡∣y∣=4(x(x2+2)342(2)3)+4(3x2+2x16+38ln⁡∣x2+2+x2∣)+xx2+2+2ln⁡∣x2+2+x∣+82(x2+2)33(2)3+C1\ln\left|y\right|=4\left(\frac{x\sqrt{\left(x^2+2\right)^{3}}}{4\sqrt{2}\sqrt{\left(2\right)^{3}}}\right)+4\left(\frac{3\sqrt{x^2+2}x}{16}+\frac{3}{8}\ln\left|\frac{\sqrt{x^2+2}+x}{\sqrt{2}}\right|\right)+x\sqrt{x^2+2}+2\ln\left|\sqrt{x^2+2}+x\right|+\frac{8\sqrt{2}\sqrt{\left(x^2+2\right)^{3}}}{3\sqrt{\left(2\right)^{3}}}+C_1

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dydx =y=(x+2)2√x2+2
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