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

解

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

\frac{3}{10} ln ∣ 2 x + 1 ∣ + \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x)) + C

解答ステップ

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

以下の部分分数を得る:

\frac{x ^{2} + x + 1}{( 2 x + 1 ) ( x ^{2} + 1 )} : \frac{3}{5 ( 2 x + 1 )} + \frac{x + 2}{5 ( x ^{2} + 1 )}

= \int \frac{3}{5 ( 2 x + 1 )} + \frac{x + 2}{5 ( 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{3}{5 ( 2 x + 1 )} d x + \int \frac{x + 2}{5 ( x ^{2} + 1 )} d x

\int \frac{3}{5 ( 2 x + 1 )} d x = \frac{3}{10} ln ∣ 2 x + 1 ∣

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

= \frac{3}{10} ln ∣ 2 x + 1 ∣ + \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x))

定数を解答に追加する

= \frac{3}{10} ln ∣ 2 x + 1 ∣ + \frac{1}{5} (\frac{1}{2} ln x^{2} + 1 + 2 arctan (x)) + C