integral of z(ln(z))^2

解答

\int z (ln (z))^{2} d z

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

求解步骤

\int z (ln (z))^{2} d z

使用分布积分法

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

\int z ln (z) d z = \frac{1}{2} z^{2} ln (z) - \frac{z ^{2}}{4}

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

化简

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

解答补常数

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

\frac{\partial}{\partial y} (cos (x^{2} + y^{2}))

\int (3 x + 5) cos (\frac{x}{4}) d x

d er i v a t i v e 3 (e^{x} cos (x) - e^{x} sin (x))

d er i v a t i v e 5 e^{- 6 x}

\frac{d}{d t} (\frac{3 t ^{2} - 3}{2 t})