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人気のある微分積分 >

integral of 6/(x^3+4x^2+6x)

解

\int \frac{6}{x ^{3} + 4 x ^{2} + 6 x} d x

解

ln ∣ x ∣ - \frac{1}{2} ln \frac{1}{2} x^{2} + 2 x + 3 - 2 arctan (\frac{1}{2} (x + 2)) + C

解答ステップ

\int \frac{6}{x ^{3} + 4 x ^{2} + 6 x} d x

定数を除く:

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

= 6 \cdot \int \frac{1}{x ^{3} + 4 x ^{2} + 6 x} d x

以下の部分分数を得る:

\frac{1}{x ^{3} + 4 x ^{2} + 6 x} : \frac{1}{6 x} + \frac{- x - 4}{6 ( x ^{2} + 4 x + 6 )}

= 6 \cdot \int \frac{1}{6 x} + \frac{- x - 4}{6 ( x ^{2} + 4 x + 6 )} d x

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

= 6 (\int \frac{1}{6 x} d x + \int \frac{- x - 4}{6 ( x ^{2} + 4 x + 6 )} d x)

\int \frac{1}{6 x} d x = \frac{1}{6} ln ∣ x ∣

\int \frac{- x - 4}{6 ( x ^{2} + 4 x + 6 )} d x = \frac{1}{12} (- ln \frac{1}{2} x^{2} + 2 x + 3 - 22 arctan (\frac{1}{2} (x + 2)))

= 6 (\frac{1}{6} ln ∣ x ∣ + \frac{1}{12} (- ln \frac{1}{2} x^{2} + 2 x + 3 - 22 arctan (\frac{1}{2} (x + 2))))

簡素化

6 (\frac{1}{6} ln ∣ x ∣ + \frac{1}{12} (- ln \frac{1}{2} x^{2} + 2 x + 3 - 22 arctan (\frac{1}{2} (x + 2)))) : ln ∣ x ∣ - \frac{1}{2} ln \frac{1}{2} x^{2} + 2 x + 3 - 2 arctan (\frac{1}{2} (x + 2))

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

定数を解答に追加する

= ln ∣ x ∣ - \frac{1}{2} ln \frac{1}{2} x^{2} + 2 x + 3 - 2 arctan (\frac{1}{2} (x + 2)) + C

グラフ

人気の例

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\int 18 x^{2} 4 6 x^{3} + 5 d x

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\int \frac{1}{x ^{2} + \frac{1}{4}} d x

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\int \frac{1}{9} cos (x) d x

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\int 10 x (5 x^{2} + 9)^{7} d x

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