integral of x^2ln(x+1)

解答

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

- x^{2} ln (x + 1) - x ln (x + 1) + \frac{1}{3} ln (x + 1) (x + 1)^{3} - \frac{1}{9} (x + 1)^{3} + \frac{1}{2} (x + 1)^{2} - x - 1 + C

求解步骤

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

使用换元积分法

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

使用分布积分法

= ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - \int \frac{1}{3} (u^{2} - 3 u + 3) d u

\int \frac{1}{3} (u^{2} - 3 u + 3) d u = \frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u

= ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - (\frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u)

u = x + 1

代回

= ln (x + 1) (\frac{( x + 1 ) ^{3}}{3} - (x + 1)^{2} + x + 1) - (\frac{( x + 1 ) ^{3}}{9} - \frac{( x + 1 ) ^{2}}{2} + x + 1)

化简

ln (x + 1) (\frac{( x + 1 ) ^{3}}{3} - (x + 1)^{2} + x + 1) - (\frac{( x + 1 ) ^{3}}{9} - \frac{( x + 1 ) ^{2}}{2} + x + 1) : - x^{2} ln (x + 1) - x ln (x + 1) + \frac{1}{3} ln (x + 1) (x + 1)^{3} - \frac{1}{9} (x + 1)^{3} + \frac{1}{2} (x + 1)^{2} - x - 1

= - x^{2} ln (x + 1) - x ln (x + 1) + \frac{1}{3} ln (x + 1) (x + 1)^{3} - \frac{1}{9} (x + 1)^{3} + \frac{1}{2} (x + 1)^{2} - x - 1

解答补常数

= - x^{2} ln (x + 1) - x ln (x + 1) + \frac{1}{3} ln (x + 1) (x + 1)^{3} - \frac{1}{9} (x + 1)^{3} + \frac{1}{2} (x + 1)^{2} - x - 1 + C

\frac{d}{d x} (e^{2 x} + 4 e^{- 4 x} + 1)

\int_{1}^{2} \frac{13}{x ^{3} + 4 x} d x

t an g e n t f (x) = x x, (16, 64)

\frac{d}{d x} (e^{x} cos (3 x) - 3 e^{x} sin (3 x))

\int_{0}^{5} π (x^{2}) d x