integral of (ln(x))^{-2}

解答

\int (ln (x))^{- 2} d x

- \frac{x}{ln ( x )} + Ei (ln (x)) + C

求解步骤

\int (ln (x))^{- 2} d x

使用指数法则:

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

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

使用换元积分法

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

使用分布积分法

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

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

= - \frac{e ^{u}}{u} - (- Ei (u))

u = ln (x)

代回

= - \frac{e ^{l n (x)}}{ln ( x )} - (- Ei (ln (x)))

化简

- \frac{e ^{l n (x)}}{ln ( x )} - (- Ei (ln (x))) : - \frac{x}{ln ( x )} + Ei (ln (x))

= - \frac{x}{ln ( x )} + Ei (ln (x))

解答补常数

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

\int (x + 8) e^{x} d x

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\int cot (3 s) csc^{4} (3 s) d s