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人気のある微分積分 >

integral of 4/((x+2)(x-2))

解

\int \frac{4}{( x + 2 ) ( x - 2 )} d x

解

- ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{4}{( x + 2 ) ( x - 2 )} d x

定数を除く:

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

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

以下の部分分数を得る:

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

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

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

\int \frac{1}{4 ( x + 2 )} d x = \frac{1}{4} ln ∣ x + 2 ∣

\int \frac{1}{4 ( x - 2 )} d x = \frac{1}{4} ln ∣ x - 2 ∣

= 4 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣)

簡素化

4 (- \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{4} ln ∣ x - 2 ∣) : - ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣

= - ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣

定数を解答に追加する

= - ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣ + C

グラフ

人気の例

integral from 0 to 1 of (x-1)/(x^2+3x+2)

\int_{0}^{1} \frac{x - 1}{x ^{2} + 3 x + 2} d x

tangent f(x)=(e^x)/x ,\at x=1

t an g e n t f (x) = \frac{e ^{x}}{x}, at x = 1

derivative of sqrt((3x+2/(5-4x)))

\frac{d}{d x} (\frac{3 x + 2}{5 - 4 x})

(\partial)/(\partial x)(1/(ln(y/x)))

\frac{\partial}{\partial x} (\frac{1}{ln ( \frac{y}{x} )})

簡約 ln(((x+6)^4e^{5x^3-3})/(x+2x^3))

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