integral from 1 to 2 of (2x+1)ln(x)

解答

\int_{1}^{2} (2 x + 1) ln (x) d x

6 ln (2) - \frac{5}{2}

十进制

1.65888 \dots

求解步骤

\int_{1}^{2} (2 x + 1) ln (x) d x

使用分布积分法

= [ln (x) (x^{2} + x) - \int x + 1 d x]_{1}^{2}

\int x + 1 d x = \frac{x ^{2}}{2} + x

= [ln (x) (x^{2} + x) - (\frac{x ^{2}}{2} + x)]_{1}^{2}

乘开

ln (x) (x^{2} + x) - (\frac{x ^{2}}{2} + x) : x^{2} ln (x) + x ln (x) - \frac{x ^{2}}{2} - x

= [x^{2} ln (x) + x ln (x) - \frac{x ^{2}}{2} - x]_{1}^{2}

带入上下限计算后得:

6 ln (2) - \frac{5}{2}

= 6 ln (2) - \frac{5}{2}

\int \frac{x ^{2} + 1}{x ^{3} + 3 x - 16} d x

\int e^{2 t} + e^{- 2 t} + 2 d t

\int x ln (x^{2} + 1) d x

in v erse l a pl a ce \frac{1}{s ^{2} ( s + 1.1 )}

\int \frac{1}{2 + x - x ^{2}} d x