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人気のある微分積分 >

integral of 1/(-x(x^2-4))

解

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

解

- \frac{1}{2} (- \frac{1}{2} ln (x) + \frac{1}{4} ln x^{2} - 4) + C

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

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

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{1}{u ( u - 4 )} d u

以下の部分分数を得る:

\frac{1}{u ( u - 4 )} : - \frac{1}{4 u} + \frac{1}{4 ( u - 4 )}

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

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

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

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

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

代用を戻す

u = x^{2}

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

\frac{1}{4} ln x^{2} = \frac{1}{2} ln (x)

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

定数を解答に追加する

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

グラフ

人気の例

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\int x^{2} (5 x + 8)^{\frac{1}{2}} d x

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\int csc^{2} (2 θ + 9) d θ

integral of 2+y

\int 2 + y d y