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人気のある微分積分 >

integral of ((2-x))/(sqrt(2x^2+5x+3))

解

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

解

\frac{1}{4 2} (- 16 x^{2} + 40 x + 24 + 13 ln 16 x^{2} + 40 x + 24 + 4 x + 5) + C

解答ステップ

\int \frac{2 - x}{2 x ^{2} + 5 x + 3} d x

平方完成する

2 x^{2} + 5 x + 3 : 2 (x + \frac{5}{4})^{2} - \frac{1}{8}

= \int \frac{2 - x}{2 ( x + \frac{5}{4} ) ^{2} - \frac{1}{8}} d x

u置換積分法を適用する

= \int \frac{- 4 u + 13}{2 16 u ^{2} - 1} d u

定数を除く:

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

= \frac{1}{2} \cdot \int \frac{- 4 u + 13}{16 u ^{2} - 1} d u

拡張

\frac{- 4 u + 13}{16 u ^{2} - 1} : - \frac{4 u}{16 u ^{2} - 1} + \frac{13}{16 u ^{2} - 1}

= \frac{1}{2} \cdot \int - \frac{4 u}{16 u ^{2} - 1} + \frac{13}{16 u ^{2} - 1} 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

= \frac{1}{2} (- \int \frac{4 u}{16 u ^{2} - 1} d u + \int \frac{13}{16 u ^{2} - 1} d u)

\int \frac{4 u}{16 u ^{2} - 1} d u = \frac{1}{4} 16 u^{2} - 1

\int \frac{13}{16 u ^{2} - 1} d u = \frac{13}{4} ln 16 u^{2} - 1 + 4 u

= \frac{1}{2} (- \frac{1}{4} 16 u^{2} - 1 + \frac{13}{4} ln 16 u^{2} - 1 + 4 u)

代用を戻す

u = x + \frac{5}{4}

= \frac{1}{2} - \frac{1}{4} 16 (x + \frac{5}{4})^{2} - 1 + \frac{13}{4} ln 16 (x + \frac{5}{4})^{2} - 1 + 4 (x + \frac{5}{4})

簡素化

\frac{1}{2} - \frac{1}{4} 16 (x + \frac{5}{4})^{2} - 1 + \frac{13}{4} ln 16 (x + \frac{5}{4})^{2} - 1 + 4 (x + \frac{5}{4}) : \frac{1}{4 2} (- 16 x^{2} + 40 x + 24 + 13 ln 16 x^{2} + 40 x + 24 + 4 x + 5)

= \frac{1}{4 2} (- 16 x^{2} + 40 x + 24 + 13 ln 16 x^{2} + 40 x + 24 + 4 x + 5)

定数を解答に追加する

= \frac{1}{4 2} (- 16 x^{2} + 40 x + 24 + 13 ln 16 x^{2} + 40 x + 24 + 4 x + 5) + C

グラフ

人気の例

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