integral of ln(a+bx)

解答

\int ln (a + b x) d x

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

求解步骤

\int ln (a + b x) d x

使用换元积分法

= \int \frac{ln ( u )}{b} d u

提出常数:

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

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

使用分布积分法

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

\int 1 d u = u

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

u = a + b x

代回

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

化简

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

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

解答补常数

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

(x^{2} + 2) y^{^{'}} + 2 x y = x^{3}

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

\int \frac{1}{3 - 6 x - x ^{2}} d x

x \to 0 lim (5 x + 4)

\frac{\partial}{\partial x} (ln (5 x^{2} + 4 y^{2} + 7))