integral of (1-z)/(-1-z)

解答

\int \frac{1 - z}{- 1 - z}

- 2 ln ∣ z + 1 ∣ - \frac{z ^{2}}{- z - 1} + \frac{1}{- z - 1} + C

求解步骤

\int \frac{1 - z}{- 1 - z} d z

乘开

\frac{1 - z}{- 1 - z} : \frac{1}{- 1 - z} - \frac{z}{- 1 - z}

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

= \int \frac{1}{- 1 - z} d z - \int \frac{z}{- 1 - z} d z

\int \frac{1}{- 1 - z} d z = - ln ∣ 1 + z ∣

\int \frac{z}{- 1 - z} d z = \frac{z ^{2}}{- 1 - z} + ln ∣ 1 + z ∣ - \frac{1}{- 1 - z}

= - ln ∣ 1 + z ∣ - (\frac{z ^{2}}{- 1 - z} + ln ∣ 1 + z ∣ - \frac{1}{- 1 - z})

化简

= - 2 ln ∣ z + 1 ∣ - \frac{z ^{2}}{- z - 1} + \frac{1}{- z - 1}

解答补常数

= - 2 ln ∣ z + 1 ∣ - \frac{z ^{2}}{- z - 1} + \frac{1}{- z - 1} + C

(ln (x))^{^{'}}

\int (x^{2} - 2 x - 3) d x

(x^{2} + 1) y^{^{'}} + y^{2} + 1 = 0, y (4) = e

\int \frac{1}{( sin ( x ) ) ^{2} ( cos ( x ) ) ^{4}} d x

\frac{d ^{4}}{d x ^{4}} (ln (x))