integral of (x+2)ln(x+2)2

解答

\int (x + 2) ln (x + 2) 2 d x

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

求解步骤

\int (x + 2) ln (x + 2) \cdot 2 d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int (x + 2) ln (x + 2) d x

乘开

(x + 2) ln (x + 2) : x ln (x + 2) + 2 ln (x + 2)

= 2 \cdot \int x ln (x + 2) + 2 ln (x + 2) d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int x ln (x + 2) d x + \int 2 ln (x + 2) d x)

\int x ln (x + 2) d x = ln (x + 2) (\frac{1}{2} (x + 2)^{2} - 2 (x + 2)) - \frac{1}{4} (x^{2} - 4 x - 12)

\int 2 ln (x + 2) d x = 2 (x ln (x + 2) + 2 ln (x + 2) - x - 2)

= 2 (ln (x + 2) (\frac{1}{2} (x + 2)^{2} - 2 (x + 2)) - \frac{1}{4} (x^{2} - 4 x - 12) + 2 (x ln (x + 2) + 2 ln (x + 2) - x - 2))

化简

2 (ln (x + 2) (\frac{1}{2} (x + 2)^{2} - 2 (x + 2)) - \frac{1}{4} (x^{2} - 4 x - 12) + 2 (x ln (x + 2) + 2 ln (x + 2) - x - 2)) : x^{2} ln (x + 2) - \frac{1}{2} x^{2} + 4 x ln (x + 2) - 2 x + 4 ln (x + 2) - 2

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

解答补常数

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

\int_{0}^{4} 5 x - x^{2} d x

\int e^{u} u d u

\int \frac{1}{29 - x ^{2}} d x

\frac{d}{d x} (ln (x + 9 + x^{2}))

a re a 2 x + 1, x^{2} - 4 x