integral of (x+1)ln(x)

解答

\int (x + 1) ln (x) d x

\frac{1}{2} x^{2} ln (x) + x ln (x) - \frac{x ^{2}}{4} - x + C

求解步骤

\int (x + 1) ln (x) d x

使用分布积分法

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

\int \frac{1}{2} (x + 2) d x = \frac{x ^{2}}{4} + x

= ln (x) (\frac{x ^{2}}{2} + x) - (\frac{x ^{2}}{4} + x)

化简

ln (x) (\frac{x ^{2}}{2} + x) - (\frac{x ^{2}}{4} + x) : \frac{1}{2} x^{2} ln (x) + x ln (x) - \frac{x ^{2}}{4} - x

= \frac{1}{2} x^{2} ln (x) + x ln (x) - \frac{x ^{2}}{4} - x

解答补常数

= \frac{1}{2} x^{2} ln (x) + x ln (x) - \frac{x ^{2}}{4} - x + C

t an g e n t f (x) = x^{2} x^{3} + 1

\int x \cdot f (x) d x

\frac{d}{d x} (2 y^{2} + 3 x)

\frac{\partial}{\partial y} (x^{3} - 6 x y + 8 y^{3})

\frac{\partial}{\partial x} (cos (7 x + 4 y + z))