integral of (x+1)ln(x+1)

解答

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

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

求解步骤

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

使用换元积分法

= \int u ln (u) d u

使用分布积分法

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

\int \frac{u}{2} d u = \frac{u ^{2}}{4}

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

u = x + 1

代回

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

化简

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

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

解答补常数

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

\int (e^{3 x} + 2)^{2} d x

\int \frac{1}{x + 5 x + 6} d x

\frac{d}{d x} (sin (\frac{π}{6}))

d er i v a t i v e f (θ) = 5 θ

\int (14 x + 15 x) d x