integral of w/(w^3+w^2+4w+4)

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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{1}{5 ( w + 1 )} d w + \int \frac{w + 4}{5 ( w ^{2} + 4 )} d w

\int \frac{1}{5 ( w + 1 )} d w = \frac{1}{5} ln ∣ w + 1 ∣

\int \frac{w + 4}{5 ( w ^{2} + 4 )} d w = \frac{1}{5} (\frac{1}{2} ln \frac{w ^{2}}{4} + 1 + 2 arctan (\frac{w}{2}))

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

定数を解答に追加する

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