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人気のある微分積分 >

integral of (x^3+1)/(x^2+5x+6)

解

\int \frac{x ^{3} + 1}{x ^{2} + 5 x + 6} d x

解

\frac{1}{8} (2 x + 5)^{2} - \frac{15}{2} (x + \frac{5}{2}) + 4 (\frac{19}{8} ln 4 x^{2} + 20 x + 24 + \frac{33}{8} ln ∣ 2 x + 6 ∣ - \frac{33}{8} ln ∣ 2 x + 4 ∣) + C

解答ステップ

\int \frac{x ^{3} + 1}{x ^{2} + 5 x + 6} d x

長除法

\frac{x ^{3} + 1}{x ^{2} + 5 x + 6} : x - 5 + \frac{19 x + 31}{x ^{2} + 5 x + 6}

= \int x - 5 + \frac{19 x + 31}{x ^{2} + 5 x + 6} d x

平方完成する

x^{2} + 5 x + 6 : (x + \frac{5}{2})^{2} - \frac{1}{4}

= \int x - 5 + \frac{19 x + 31}{( x + \frac{5}{2} ) ^{2} - \frac{1}{4}} d x

u置換積分法を適用する

= \int u - \frac{5}{2} - 5 + \frac{4 ( 19 ( u - \frac{5}{2} ) + 31 )}{4 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

= \int u d u - \int \frac{5}{2} d u - \int 5 d u + \int \frac{4 ( 19 ( u - \frac{5}{2} ) + 31 )}{4 u ^{2} - 1} d u

\int u d u = \frac{u ^{2}}{2}

\int \frac{5}{2} d u = \frac{5}{2} u

\int 5 d u = 5 u

\int \frac{4 ( 19 ( u - \frac{5}{2} ) + 31 )}{4 u ^{2} - 1} d u = 4 (\frac{19}{8} ln 4 u^{2} - 1 + \frac{33}{8} ln ∣ 2 u + 1 ∣ - \frac{33}{8} ln ∣ 2 u - 1 ∣)

= \frac{u ^{2}}{2} - \frac{5}{2} u - 5 u + 4 (\frac{19}{8} ln 4 u^{2} - 1 + \frac{33}{8} ln ∣ 2 u + 1 ∣ - \frac{33}{8} ln ∣ 2 u - 1 ∣)

代用を戻す

u = x + \frac{5}{2}

= \frac{( x + \frac{5}{2} ) ^{2}}{2} - \frac{5}{2} (x + \frac{5}{2}) - 5 (x + \frac{5}{2}) + 4 (\frac{19}{8} ln 4 (x + \frac{5}{2})^{2} - 1 + \frac{33}{8} ln 2 (x + \frac{5}{2}) + 1 - \frac{33}{8} ln 2 (x + \frac{5}{2}) - 1)

簡素化

\frac{( x + \frac{5}{2} ) ^{2}}{2} - \frac{5}{2} (x + \frac{5}{2}) - 5 (x + \frac{5}{2}) + 4 (\frac{19}{8} ln 4 (x + \frac{5}{2})^{2} - 1 + \frac{33}{8} ln 2 (x + \frac{5}{2}) + 1 - \frac{33}{8} ln 2 (x + \frac{5}{2}) - 1) : \frac{1}{8} (2 x + 5)^{2} - \frac{15}{2} (x + \frac{5}{2}) + 4 (\frac{19}{8} ln 4 x^{2} + 20 x + 24 + \frac{33}{8} ln ∣ 2 x + 6 ∣ - \frac{33}{8} ln ∣ 2 x + 4 ∣)

= \frac{1}{8} (2 x + 5)^{2} - \frac{15}{2} (x + \frac{5}{2}) + 4 (\frac{19}{8} ln 4 x^{2} + 20 x + 24 + \frac{33}{8} ln ∣ 2 x + 6 ∣ - \frac{33}{8} ln ∣ 2 x + 4 ∣)

定数を解答に追加する

= \frac{1}{8} (2 x + 5)^{2} - \frac{15}{2} (x + \frac{5}{2}) + 4 (\frac{19}{8} ln 4 x^{2} + 20 x + 24 + \frac{33}{8} ln ∣ 2 x + 6 ∣ - \frac{33}{8} ln ∣ 2 x + 4 ∣) + C

グラフ

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