integral of ln(5x+1)

解答

\int ln (5 x + 1) d x

x ln (5 x + 1) + \frac{1}{5} ln (5 x + 1) - x - \frac{1}{5} + C

求解步骤

\int ln (5 x + 1) d x

使用换元积分法

= \int ln (u) \frac{1}{5} d u

提出常数:

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

= \frac{1}{5} \cdot \int ln (u) d u

使用分布积分法

= \frac{1}{5} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{5} (u ln (u) - u)

u = 5 x + 1

代回

= \frac{1}{5} ((5 x + 1) ln (5 x + 1) - (5 x + 1))

化简

\frac{1}{5} ((5 x + 1) ln (5 x + 1) - (5 x + 1)) : x ln (5 x + 1) + \frac{1}{5} ln (5 x + 1) - x - \frac{1}{5}

= x ln (5 x + 1) + \frac{1}{5} ln (5 x + 1) - x - \frac{1}{5}

解答补常数

= x ln (5 x + 1) + \frac{1}{5} ln (5 x + 1) - x - \frac{1}{5} + C

\int \frac{5 - 4 x}{4 - 4 x} d x

d er i v a t i v e y = x^{3} e^{- 8 x}

\frac{\partial}{\partial y} (\frac{9 y}{x ^{3}})

\int 0.5 e^{- x} d x

\int \frac{4}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x