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人気のある微分積分 >

integral of (-16x+3)/((x^2+2)^2)

解

\int \frac{- 16 x + 3}{( x ^{2} + 2 ) ^{2}} d x

解

2 (\frac{4 2}{x ^{2} + 2} + \frac{3}{8} (arctan (\frac{x}{2}) + \frac{2 x}{2 + x ^{2}})) + C

解答ステップ

\int \frac{- 16 x + 3}{( x ^{2} + 2 ) ^{2}} d x

置換積分法を適用する

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

拡張

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

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

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

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

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

代用を戻す

u = \frac{x}{2}

= 2 - - \frac{2 2}{( \frac{x}{2} ) ^{2} + 1} + \frac{3}{8} arctan (\frac{x}{2}) + \frac{\frac{x}{2}}{1 + ( \frac{x}{2} ) ^{2}}

簡素化

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

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

定数を解答に追加する

= 2 (\frac{4 2}{x ^{2} + 2} + \frac{3}{8} (arctan (\frac{x}{2}) + \frac{2 x}{2 + x ^{2}})) + C

グラフ

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