integral of (ln(y))^2y

解答

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

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

求解步骤

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

使用分布积分法

: \frac{1}{2} y^{2} ln^{2} (y) - \int y ln (y) d y

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

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

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

化简

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

解答补常数

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

\frac{x ^{2} + 4}{x ^{2} - 4}

6 - 6 x^{2}, 0

\frac{d}{d x} (- \frac{3}{8 x ^{\frac{5}{2}}})

\int sec^{2} (5 x) + e^{\frac{x}{2}} d x

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