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人気のある微分積分 >

integral of (x+2)/(x^2+7x+12)

解

\int \frac{x + 2}{x ^{2} + 7 x + 12} d x

解

2 (\frac{1}{4} ln 4 x^{2} + 28 x + 48 + \frac{3}{4} (ln ∣ 2 x + 8 ∣ - ln ∣ 2 x + 6 ∣)) + C

解答ステップ

\int \frac{x + 2}{x ^{2} + 7 x + 12} d x

平方完成する

x^{2} + 7 x + 12 : (x + \frac{7}{2})^{2} - \frac{1}{4}

= \int \frac{x + 2}{( x + \frac{7}{2} ) ^{2} - \frac{1}{4}} d x

u置換積分法を適用する

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

拡張

\frac{2 u - 3}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} - \frac{3}{4 u ^{2} - 1}

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{2 u}{4 u ^{2} - 1} d u - \int \frac{3}{4 u ^{2} - 1} d u)

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

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

= 2 (\frac{1}{4} ln 4 u^{2} - 1 - (- \frac{3}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)))

代用を戻す

u = x + \frac{7}{2}

= 2 (\frac{1}{4} ln 4 (x + \frac{7}{2})^{2} - 1 - (- \frac{3}{4} (ln 2 (x + \frac{7}{2}) + 1 - ln 2 (x + \frac{7}{2}) - 1)))

簡素化

2 (\frac{1}{4} ln 4 (x + \frac{7}{2})^{2} - 1 - (- \frac{3}{4} (ln 2 (x + \frac{7}{2}) + 1 - ln 2 (x + \frac{7}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 28 x + 48 + \frac{3}{4} (ln ∣ 2 x + 8 ∣ - ln ∣ 2 x + 6 ∣))

= 2 (\frac{1}{4} ln 4 x^{2} + 28 x + 48 + \frac{3}{4} (ln ∣ 2 x + 8 ∣ - ln ∣ 2 x + 6 ∣))

定数を解答に追加する

= 2 (\frac{1}{4} ln 4 x^{2} + 28 x + 48 + \frac{3}{4} (ln ∣ 2 x + 8 ∣ - ln ∣ 2 x + 6 ∣)) + C

グラフ

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