integral of 3/((z^2+4)(z+1)^2)

解

\int \frac{3}{( z ^{2} + 4 ) ( z + 1 ) ^{2}} d z

3 (\frac{1}{25} (- ln z^{2} + 4 - \frac{3}{2} arctan (\frac{z}{2})) + \frac{2}{25} ln ∣ z + 1 ∣ - \frac{1}{5 ( z + 1 )}) + C

解答ステップ

\int \frac{3}{( z ^{2} + 4 ) ( z + 1 ) ^{2}} d z

定数を除く:

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

= 3 \cdot \int \frac{1}{( z ^{2} + 4 ) ( z + 1 ) ^{2}} d z

以下の部分分数を得る:

\frac{1}{( z ^{2} + 4 ) ( z + 1 ) ^{2}} : \frac{- 2 z - 3}{25 ( z ^{2} + 4 )} + \frac{2}{25 ( z + 1 )} + \frac{1}{5 ( z + 1 ) ^{2}}

= 3 \cdot \int \frac{- 2 z - 3}{25 ( z ^{2} + 4 )} + \frac{2}{25 ( z + 1 )} + \frac{1}{5 ( z + 1 ) ^{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

= 3 (\int \frac{- 2 z - 3}{25 ( z ^{2} + 4 )} d z + \int \frac{2}{25 ( z + 1 )} d z + \int \frac{1}{5 ( z + 1 ) ^{2}} d z)

\int \frac{- 2 z - 3}{25 ( z ^{2} + 4 )} d z = \frac{1}{25} (- ln z^{2} + 4 - \frac{3}{2} arctan (\frac{z}{2}))

\int \frac{2}{25 ( z + 1 )} d z = \frac{2}{25} ln ∣ z + 1 ∣

\int \frac{1}{5 ( z + 1 ) ^{2}} d z = - \frac{1}{5 ( z + 1 )}

= 3 (\frac{1}{25} (- ln z^{2} + 4 - \frac{3}{2} arctan (\frac{z}{2})) + \frac{2}{25} ln ∣ z + 1 ∣ - \frac{1}{5 ( z + 1 )})

定数を解答に追加する

= 3 (\frac{1}{25} (- ln z^{2} + 4 - \frac{3}{2} arctan (\frac{z}{2})) + \frac{2}{25} ln ∣ z + 1 ∣ - \frac{1}{5 ( z + 1 )}) + C