integral of (8x^2+8x+2)/((4x^2+1)^2)

解

\int \frac{8 x ^{2} + 8 x + 2}{( 4 x ^{2} + 1 ) ^{2}} d x

arctan (2 x) - \frac{1}{4 x ^{2} + 1} + C

解答ステップ

\int \frac{8 x ^{2} + 8 x + 2}{( 4 x ^{2} + 1 ) ^{2}} d x

以下の部分分数を得る:

\frac{8 x ^{2} + 8 x + 2}{( 4 x ^{2} + 1 ) ^{2}} : \frac{2}{4 x ^{2} + 1} + \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}}

= \int \frac{2}{4 x ^{2} + 1} + \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}} 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{2}{4 x ^{2} + 1} d x + \int \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}} d x

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

\int \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}} d x = - \frac{1}{4 x ^{2} + 1}

= arctan (2 x) - \frac{1}{4 x ^{2} + 1}

定数を解答に追加する

= arctan (2 x) - \frac{1}{4 x ^{2} + 1} + C