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受欢迎的微积分 >

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

解答

\int \frac{1}{1 + ln ( x )} d x

解答

\frac{1}{e} Ei (1 + ln (x)) + C

求解步骤

\int \frac{1}{1 + ln ( x )} d x

使用换元积分法

= \int \frac{e ^{u - 1}}{u} d u

化简

e^{u - 1} : e^{u} e^{- 1}

= \int \frac{e ^{u} e ^{- 1}}{u} d u

提出常数:

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

= e^{- 1} \cdot \int \frac{e ^{u}}{u} d u

使用指数法则:

a^{- 1} = \frac{1}{a}

= \frac{1}{e} \cdot \int \frac{e ^{u}}{u} d u

这是非初等函数积分:

\int \frac{e ^{x}}{x} d x = Ei (x)

= \frac{1}{e} Ei (u)

u = 1 + ln (x)

代回

= \frac{1}{e} Ei (1 + ln (x))

解答补常数

= \frac{1}{e} Ei (1 + ln (x)) + C

流行的例子

tangent f(x)=x^5,\at x=-1

t an g e n t f (x) = x^{5}, at x = - 1

integral of x/((1-x)^2)

\int \frac{x}{( 1 - x ) ^{2}} d x

limit as x approaches 0 of (9x^2-8x)/x

x \to 0 lim (\frac{9 x ^{2} - 8 x}{x})

- y + y^{^{'}} = x + e^{2} x

y^'+2/x y=x^2y^2

y^{^{'}} + \frac{2}{x} y = x^{2} y^{2}