The value of the integral _0^2x(x^2+x+1)(x+1)(x^4+x^2+1)dx is equal to:

Question

The value of the integral $\int _{0}^{2}\frac{\sqrt{x({x}^{2}+x+1)}}{(\sqrt{x+1})(\sqrt{{x}^{4}+{x}^{2}+1})}dx$ is equal to:

Accepted Answer

$I=\int _{0}^{2}[\frac{\sqrt{x({x}^{2}+x+1)}}{(\sqrt{x+1})\cdot (\sqrt{{x}^{4}+{x}^{2}+1})}]dx$

$\begin{gathered} ={\int }_{2}^{0}[[|\frac{\sqrt{x({x}^{2}+x+1)}}{(\sqrt{x+1})\cdot (\sqrt{{x}^{4}+2{x}^{2}+1-{x}^{2}})}]]|dx \ ={\int }_{2}^{0}[[|\frac{\sqrt{x({x}^{2}+x+1)}}{(\sqrt{x+1})\cdot (\sqrt{{({x}^{2}+1)}^{2}-{x}^{2}})}]]|dx \ ={\int }_{2}^{0}[[|\frac{\sqrt{x({x}^{2}+x+1)}}{(\sqrt{x+1})\cdot {\sqrt{({x}^{2}+1+x)({x}^{2}+1-x)}}]}]]|dx \ ={\int }_{2}^{0}[[|\frac{(\sqrt{x})\cdot (\sqrt{{x}^{2}+x+1})}{(\sqrt{x+1})\cdot (\sqrt{{x}^{2}+1+x})\cdot (\sqrt{{x}^{2}+1-x})}]]|dx \ ={\int }_{2}^{0}[[|\frac{\sqrt{x}}{(\sqrt{x+1})\cdot (\sqrt{{x}^{2}-x+1})}]]|dx \ ={\int }_{2}^{0}[\frac{\sqrt{x}}{\sqrt{(x+1)({x}^{2}-x+1)}}]dx \ ={\int }_{2}^{0}[\frac{\sqrt{x}}{\sqrt{{x}^{3}+1}}]dx \end{gathered}$

Let $x={t}^{2}\Rightarrow dx=2tdt$

$I={\int }_{\sqrt{2}}^{0}[\frac{2{t}^{2}}{\sqrt{{t}^{6}+1}}]dt=(\frac{2}{3}){\int }_{\sqrt{2}}^{0}[\frac{3{t}^{2}}{\sqrt{{t}^{6}+1}}]dt$

Let ${t}^{3}=u\Rightarrow 3{t}^{2}dt=du$

$I={(\frac{2}{3})}^{2}\cdot {\int }_{2\sqrt{2}}^{0}[\frac{du}{\sqrt{{u}^{2}+1}}]$

Using standard integration formulaes :

$\int (\frac{1}{\sqrt{{x}^{2}+{a}^{2}}})dx=\ln |x+\sqrt{{x}^{2}+{a}^{2}}|+C$

= $(\frac{2}{3})\cdot {[\ln |u+\sqrt{{u}^{2}+1}|]}_{2\sqrt{2}}^{0}$

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

Solutions · 1