integral of (10z+2)/((z+1)^2(z-3)^2)

解

\int \frac{10 z + 2}{( z + 1 ) ^{2} ( z - 3 ) ^{2}} d z

\frac{3}{8} ln ∣ z + 1 ∣ + \frac{1}{2 ( z + 1 )} - \frac{3}{8} ln ∣ z - 3 ∣ - \frac{2}{z - 3} + C

解答ステップ

\int \frac{10 z + 2}{( z + 1 ) ^{2} ( z - 3 ) ^{2}} d z

以下の部分分数を得る:

\frac{10 z + 2}{( z + 1 ) ^{2} ( z - 3 ) ^{2}} : \frac{3}{8 ( z + 1 )} - \frac{1}{2 ( z + 1 ) ^{2}} - \frac{3}{8 ( z - 3 )} + \frac{2}{( z - 3 ) ^{2}}

= \int \frac{3}{8 ( z + 1 )} - \frac{1}{2 ( z + 1 ) ^{2}} - \frac{3}{8 ( z - 3 )} + \frac{2}{( z - 3 ) ^{2}} d z

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3}{8 ( z + 1 )} d z - \int \frac{1}{2 ( z + 1 ) ^{2}} d z - \int \frac{3}{8 ( z - 3 )} d z + \int \frac{2}{( z - 3 ) ^{2}} d z

\int \frac{3}{8 ( z + 1 )} d z = \frac{3}{8} ln ∣ z + 1 ∣

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

\int \frac{3}{8 ( z - 3 )} d z = \frac{3}{8} ln ∣ z - 3 ∣

\int \frac{2}{( z - 3 ) ^{2}} d z = - \frac{2}{z - 3}

= \frac{3}{8} ln ∣ z + 1 ∣ - (- \frac{1}{2 ( z + 1 )}) - \frac{3}{8} ln ∣ z - 3 ∣ - \frac{2}{z - 3}

簡素化

= \frac{3}{8} ln ∣ z + 1 ∣ + \frac{1}{2 ( z + 1 )} - \frac{3}{8} ln ∣ z - 3 ∣ - \frac{2}{z - 3}

定数を解答に追加する

= \frac{3}{8} ln ∣ z + 1 ∣ + \frac{1}{2 ( z + 1 )} - \frac{3}{8} ln ∣ z - 3 ∣ - \frac{2}{z - 3} + C