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人気のある微分積分 >

integral of ((8x^3+3))/((x^2+2)^2)

解

\int \frac{( 8 x ^{3} + 3 )}{( x ^{2} + 2 ) ^{2}} d x

解

2 (22 ln \frac{x ^{2}}{2} + 1 + \frac{1}{4} (\frac{16 2}{x ^{2} + 2} + \frac{3}{2} (arctan (\frac{x}{2}) + \frac{2 x}{2 + x ^{2}}))) + C

解答ステップ

\int \frac{8 x ^{3} + 3}{( x ^{2} + 2 ) ^{2}} d x

置換積分法を適用する

= \int \frac{2 ( 16 2 u ^{3} + 3 )}{( 2 u ^{2} + 2 ) ^{2}} d u

定数を除く:

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

= 2 \cdot \int \frac{16 2 u ^{3} + 3}{( 2 u ^{2} + 2 ) ^{2}} d u

以下の部分分数を得る:

\frac{16 2 u ^{3} + 3}{( 2 u ^{2} + 2 ) ^{2}} : \frac{4 2 u}{u ^{2} + 1} + \frac{- 16 2 u + 3}{4 ( u ^{2} + 1 ) ^{2}}

= 2 \cdot \int \frac{4 2 u}{u ^{2} + 1} + \frac{- 16 2 u + 3}{4 ( u ^{2} + 1 ) ^{2}} d u

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

= 2 (\int \frac{4 2 u}{u ^{2} + 1} d u + \int \frac{- 16 2 u + 3}{4 ( u ^{2} + 1 ) ^{2}} d u)

\int \frac{4 2 u}{u ^{2} + 1} d u = 22 ln u^{2} + 1

\int \frac{- 16 2 u + 3}{4 ( u ^{2} + 1 ) ^{2}} d u = \frac{1}{4} (\frac{8 2}{u ^{2} + 1} + \frac{3}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

= 2 (22 ln u^{2} + 1 + \frac{1}{4} (\frac{8 2}{u ^{2} + 1} + \frac{3}{2} (arctan (u) + \frac{u}{1 + u ^{2}})))

代用を戻す

u = \frac{x}{2}

= 2 22 ln (\frac{x}{2})^{2} + 1 + \frac{1}{4} \frac{8 2}{( \frac{x}{2} ) ^{2} + 1} + \frac{3}{2} arctan (\frac{x}{2}) + \frac{\frac{x}{2}}{1 + ( \frac{x}{2} ) ^{2}}

簡素化

2 22 ln (\frac{x}{2})^{2} + 1 + \frac{1}{4} \frac{8 2}{( \frac{x}{2} ) ^{2} + 1} + \frac{3}{2} arctan (\frac{x}{2}) + \frac{\frac{x}{2}}{1 + ( \frac{x}{2} ) ^{2}} : 2 (22 ln \frac{x ^{2}}{2} + 1 + \frac{1}{4} (\frac{16 2}{x ^{2} + 2} + \frac{3}{2} (arctan (\frac{x}{2}) + \frac{2 x}{2 + x ^{2}})))

= 2 (22 ln \frac{x ^{2}}{2} + 1 + \frac{1}{4} (\frac{16 2}{x ^{2} + 2} + \frac{3}{2} (arctan (\frac{x}{2}) + \frac{2 x}{2 + x ^{2}})))

定数を解答に追加する

= 2 (22 ln \frac{x ^{2}}{2} + 1 + \frac{1}{4} (\frac{16 2}{x ^{2} + 2} + \frac{3}{2} (arctan (\frac{x}{2}) + \frac{2 x}{2 + x ^{2}}))) + C

グラフ

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