$${\int\limits_{0}^{1}{ {{x}^{2}} \ln(1+x)}{d}{x}}$$
Рішення:
$${\int\limits_{0}^{1}{ {{x}^{2}} \ln(1+x)}{d}{x}}= {\int\limits_{0}^{1}{\ln(1+x)}{d}\left({\frac{1}{3} {{x}^{3}}}\right)} = -{\int\limits_{0}^{1}{\frac{1}{3} {{x}^{3}}}{d}\left({\ln(1+x)}\right)}+{\left.{\frac{1}{3} {{x}^{3}} \ln(1+x)}\right|_{0}^{1}} = $$
$$ -{\int\limits_{0}^{1}{\frac{1}{3} \frac{{{x}^{3}}}{1+x}}{d}{x}}+{\left.{\frac{1}{3} {{x}^{3}} \ln(1+x)}\right|_{0}^{1}} = -{\int\limits_{0}^{1}{\frac{1}{3} \frac{{{x}^{3}}}{1+x}}{d}{x}}+\frac{1}{3} \ln(2) = -{\int\limits_{0}^{1}\left({\frac{1}{3}-\frac{1}{3} \frac{1}{1+x}+\frac{1}{3} {{x}^{2}}-\frac{1}{3} x}\right){d}{x}}+\frac{1}{3} \ln(2) = \frac{1}{3} \ln(2)-{\int\limits_{0}^{1}\left({\frac{1}{3}+\frac{1}{3} {{x}^{2}}-\frac{1}{3} x}\right){d}{x}}-{\int\limits_{0}^{1}\left({-\frac{1}{3} \frac{1}{1+x}}\right){d}{x}}$$
Обчислимо кожен інтеграл суми:
$${\int\limits_{0}^{1}\left({-\frac{1}{3} \frac{1}{1+x}}\right){d}{x}} = {\int\limits_{0}^{1}\left({-\frac{1}{3} \frac{1}{1+x}}\right){d}\left({1+x}\right)} = -\frac{1}{3} {\int\limits_{0}^{1}{\frac{1}{1+x}}{d}\left({1+x}\right)}=$$ $${\left.\left({-\frac{1}{3} \ln({|1+x|})}\right)\right|_{0}^{1}} = -\frac{1}{3} \ln(2) $$
$${\int\limits_{0}^{1}\left({\frac{1}{3}+\frac{1}{3} {{x}^{2}}-\frac{1}{3} x}\right){d}{x}} = {\left.\left({-\frac{1}{6} {{x}^{2}}}\right)\right|_{0}^{1}}+{\left.{\frac{1}{9} {{x}^{3}}}\right|_{0}^{1}}+{\left.{\frac{1}{3} x}\right|_{0}^{1}} = \frac{5}{18}$$
Таким чином, сума інтегралів дорівнює:
$$\frac{1}{3} \ln(2)-{\int\limits_{0}^{1}\left({\frac{1}{3}+\frac{1}{3} {{x}^{2}}-\frac{1}{3} x}\right){d}{x}}-{\int\limits_{0}^{1}\left({-\frac{1}{3} \frac{1}{1+x}}\right){d}{x}}=-\frac{5}{18}+\frac{2}{3} \ln(2)$$
$${{{\int\limits_{0}^{1}{ {{x}^{2}} \ln(1+x)}{d}{x}}}={-\frac{5}{18}+\frac{2}{3} \ln(2)}}$$