1/ [tex]\int\limits_{0}^{1}{xln(x^2+x+1)}dx[/tex]
2/ [tex]\int\limits_{0}^{3ln2}{\frac{1}{(\sqrt[3]{e^x}+2)^2}}dx[/tex]
\[\begin{array}{l}
\int_0^1 {x\ln \left( {{x^2} + x + 1} \right)dx} \\
t = {x^2} + x + 1\\
\to dt = \left( {2x + 1} \right)dx\\
\to I = \int_0^1 {\left( {\left( {x + \frac{1}{2}} \right)\ln \left( {{x^2} + x + 1} \right) - \frac{{\ln \left( {{x^2} + x + 1} \right)}}{2}} \right)dx} = \int_1^3 {\frac{{\ln t}}{2}dt} - \int_0^1 {\frac{{\ln \left( {{x^2} + x + 1} \right)}}{2}dx} = {I_1} - \frac{{{I_2}}}{2}\\
{I_1} = \int_1^3 {\frac{{\ln tdt}}{2}} = \left. {\frac{{t\ln t}}{2}} \right|_1^3 - \int_1^3 {\frac{{dt}}{2}} = ...\\
{I_2} = \int_0^1 {\ln \left( {{x^2} + x + 1} \right)dx} = \left. {x\ln \left( {{x^2} + x + 1} \right)} \right|_0^1 - \int_0^1 {\frac{{x\left( {2x + 1} \right)}}{{{x^2} + x + 1}}dx} \\
{I_3} = \int_0^1 {\frac{{x\left( {2x + 1} \right)}}{{{x^2} + x + 1}}dx} = \int_0^1 {\left( {2 - \frac{{x + \frac{1}{2}}}{{{x^2} + x + 1}} - \frac{3}{{2\left( {{x^2} + x + 1} \right)}}} \right)dx = \int_0^1 {2dx} - \int_0^1 {\frac{{x + \frac{1}{2}}}{{{x^2} + x + 1}}dx} - \frac{3}{2}\int_0^1 {\frac{1}{{{x^2} + x + 1}}dx} } \\
{I_4} = \int_0^1 {\frac{{x + \frac{1}{2}}}{{{x^2} + x + 1}}dx} \\
t = {x^2} + x + 1\\
\to {I_4} = \int_1^3 {\frac{{dt}}{{2t}}} \\
{I_5} = \int_0^1 {\frac{1}{{{x^2} + x + 1}}dx} = \int_0^1 {\frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{3}{4}}}dx} \\
x + \frac{1}{2} = \frac{{\sqrt 3 \tan t}}{2} \to dx = \frac{{\sqrt 3 }}{{2{{\cos }^2}t}}dt\\
\to {I_5} = \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sqrt 3 dt}}{{2{{\cos }^2}t\left( {{{\tan }^2}t + 1} \right)*\frac{3}{4}}}} = \frac{{2\sqrt 3 }}{3}\int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {dt} = ...
\end{array}\]
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