integral of (1-z)/(z^2+2z+1)

解答

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

- \frac{2}{z + 1} - ln ∣ z + 1 ∣ + C

求解步骤

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

乘开

\frac{1 - z}{z ^{2} + 2 z + 1} : \frac{1}{z ^{2} + 2 z + 1} - \frac{z}{z ^{2} + 2 z + 1}

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

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

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

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

化简

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

= - \frac{2}{z + 1} - ln ∣ z + 1 ∣

解答补常数

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

\int \frac{3 x}{( x + 5 ) ^{2}} d x

\int \frac{1}{6 x ^{2} - 5 x - 6} d x

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

\int t^{2} - 3 t + 2 d t

\int 2 x \cdot x^{2} d x