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

解

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

- \frac{1}{25} ln ∣ x + 2 ∣ + \frac{4}{5 ( x + 2 )} + \frac{1}{25} (\frac{1}{2} ln x^{2} + 1 + 18 arctan (x)) + C

解答ステップ

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

以下の部分分数を得る:

\frac{3 x + 2}{( x ^{2} + 4 x + 4 ) ( x ^{2} + 1 )} : - \frac{1}{25 ( x + 2 )} - \frac{4}{5 ( x + 2 ) ^{2}} + \frac{x + 18}{25 ( x ^{2} + 1 )}

= \int - \frac{1}{25 ( x + 2 )} - \frac{4}{5 ( x + 2 ) ^{2}} + \frac{x + 18}{25 ( x ^{2} + 1 )} d x

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}{25 ( x + 2 )} d x - \int \frac{4}{5 ( x + 2 ) ^{2}} d x + \int \frac{x + 18}{25 ( x ^{2} + 1 )} d x

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

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

\int \frac{x + 18}{25 ( x ^{2} + 1 )} d x = \frac{1}{25} (\frac{1}{2} ln x^{2} + 1 + 18 arctan (x))

= - \frac{1}{25} ln ∣ x + 2 ∣ - (- \frac{4}{5 ( x + 2 )}) + \frac{1}{25} (\frac{1}{2} ln x^{2} + 1 + 18 arctan (x))

簡素化

= - \frac{1}{25} ln ∣ x + 2 ∣ + \frac{4}{5 ( x + 2 )} + \frac{1}{25} (\frac{1}{2} ln x^{2} + 1 + 18 arctan (x))

定数を解答に追加する

= - \frac{1}{25} ln ∣ x + 2 ∣ + \frac{4}{5 ( x + 2 )} + \frac{1}{25} (\frac{1}{2} ln x^{2} + 1 + 18 arctan (x)) + C