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人気のある微分積分 >

integral of (x+8)/(2x^3-18x)

解

\int \frac{x + 8}{2 x ^{3} - 18 x} d x

解

\frac{1}{36} (5 ln ∣ x + 3 ∣ + 11 ln ∣ x - 3 ∣ - 16 ln ∣ x ∣) + C

解答ステップ

\int \frac{x + 8}{2 x ^{3} - 18 x} d x

因数

\frac{x + 8}{2 x ^{3} - 18 x}

= \int \frac{x + 8}{2 x ( x + 3 ) ( x - 3 )} d x

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{x + 8}{x ( x + 3 ) ( x - 3 )} d x

拡張

\frac{x + 8}{x ( x + 3 ) ( x - 3 )} : \frac{x}{x ( x + 3 ) ( x - 3 )} + \frac{8}{x ( x + 3 ) ( x - 3 )}

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{x}{x ( x + 3 ) ( x - 3 )} d x + \int \frac{8}{x ( x + 3 ) ( x - 3 )} d x)

\int \frac{x}{x ( x + 3 ) ( x - 3 )} d x = - \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x - 3 ∣

\int \frac{8}{x ( x + 3 ) ( x - 3 )} d x = 8 (\frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{18} ln ∣ x - 3 ∣)

= \frac{1}{2} (- \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x - 3 ∣ + 8 (\frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{18} ln ∣ x - 3 ∣))

簡素化

\frac{1}{2} (- \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x - 3 ∣ + 8 (\frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{18} ln ∣ x - 3 ∣)) : \frac{1}{36} (5 ln ∣ x + 3 ∣ + 11 ln ∣ x - 3 ∣ - 16 ln ∣ x ∣)

= \frac{1}{36} (5 ln ∣ x + 3 ∣ + 11 ln ∣ x - 3 ∣ - 16 ln ∣ x ∣)

定数を解答に追加する

= \frac{1}{36} (5 ln ∣ x + 3 ∣ + 11 ln ∣ x - 3 ∣ - 16 ln ∣ x ∣) + C

グラフ

人気の例

integral of ln(4)

\int ln (4)

y^{''}-6y^'+9y=0,y(0)=3,y^'(0)= 25/3

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{25}{3}

derivative of (x^4/4 ln(x)-(x^4)/(16))

\frac{d}{d x} (\frac{x ^{4}}{4} ln (x) - \frac{x ^{4}}{16})

limit as x approaches-infinity of x+x^3

x \to - \infty lim (x + x^{3})

slopeintercept (1.2)(4.4)

s l o p e in t erce pt (1.2) (4.4)