The value of the integral \({\int }_{0}^{2}\frac{\sqrt{x\left({x}^{2}+x+1\right)}}{(\sqrt{x+1})\left(\sqrt{{x}^{4}+{x}^{…

Let \(I=\int_0^2\left[\frac{\sqrt{x\left(x^2+x+1\right)}}{(\sqrt{x+1}) \cdot\left(\sqrt{x^4+x^2+1}\right)}\right] d x\)

\(=\int_0^2\left[\frac{\sqrt{x\left(x^2+x+1\right)}}{(\sqrt{x+1}) \cdot\left(\sqrt{\left(x^2+1\right)^2-x^2}\right)}\right] d x\)

\(=\int_0^2\left[\frac{\sqrt{x\left(x^2+x+1\right)}}{\left.(\sqrt{x+1}) \cdot\left\{\sqrt{\left(x^2+1+x\right)\left(x^2+1-x\right)}\right\}\right]}\right] d x\)

\(=\int_0^2\left[\frac{\sqrt{x}}{(\sqrt{x+1}) \cdot\left(\sqrt{x^2-x+1}\right)}\right] d x\)

\(=\int_0^2\left[\frac{\sqrt{x}}{\sqrt{x^3+1}}\right] d x\)

Let \(x=t^2 \Rightarrow d x=2 t d t\)

\(I=\int_0^{\sqrt{2}}\left[\frac{2 t^2}{\sqrt{t^6+1}}\right] d t=\left(\frac{2}{3}\right) \int_0^{\sqrt{2}}\left[\frac{3 t^2}{\sqrt{t^6+1}}\right] d t\)

Let \(t^3=u \Rightarrow 3 t^2 d t=d u\)

\(I=\left(\frac{2}{3}\right) \cdot \int_0^{2 \sqrt{2}}\left[\frac{d u}{\sqrt{u^2+1}}\right]\)

\(=\left(\frac{2}{3}\right) \cdot\left[\ln \left|u+\sqrt{u^2+1}\right|\right]_0^{2 \sqrt{2}}\)

\(=\left(\frac{2}{3}\right) \cdot[\ln (2 \sqrt{2}+3)-\ln 1]=\left(\frac{2}{3}\right) \cdot[\ln (3+2 \sqrt{2})]\)