ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral of 7/(x(x^4+8))

解

\int \frac{7}{x ( x ^{4} + 8 )} d x

解

\frac{7}{4} (\frac{1}{2} ln (x) - \frac{1}{8} ln x^{4} + 8) + C

解答ステップ

\int \frac{7}{x ( x ^{4} + 8 )} d x

定数を除く:

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

= 7 \cdot \int \frac{1}{x ( x ^{4} + 8 )} d x

u置換積分法を適用する

= 7 \cdot \int \frac{1}{4 u ( u + 8 )} d u

定数を除く:

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

= 7 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 8 )} d u

以下の部分分数を得る:

\frac{1}{u ( u + 8 )} : \frac{1}{8 u} - \frac{1}{8 ( u + 8 )}

= 7 \cdot \frac{1}{4} \cdot \int \frac{1}{8 u} - \frac{1}{8 ( u + 8 )} 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

= 7 \cdot \frac{1}{4} (\int \frac{1}{8 u} d u - \int \frac{1}{8 ( u + 8 )} d u)

\int \frac{1}{8 u} d u = \frac{1}{8} ln ∣ u ∣

\int \frac{1}{8 ( u + 8 )} d u = \frac{1}{8} ln ∣ u + 8 ∣

= 7 \cdot \frac{1}{4} (\frac{1}{8} ln ∣ u ∣ - \frac{1}{8} ln ∣ u + 8 ∣)

代用を戻す

u = x^{4}

= 7 \cdot \frac{1}{4} (\frac{1}{8} ln x^{4} - \frac{1}{8} ln x^{4} + 8)

簡素化

7 \cdot \frac{1}{4} (\frac{1}{8} ln x^{4} - \frac{1}{8} ln x^{4} + 8) : \frac{7}{4} (\frac{1}{2} ln (x) - \frac{1}{8} ln x^{4} + 8)

= \frac{7}{4} (\frac{1}{2} ln (x) - \frac{1}{8} ln x^{4} + 8)

定数を解答に追加する

= \frac{7}{4} (\frac{1}{2} ln (x) - \frac{1}{8} ln x^{4} + 8) + C

グラフ

人気の例

derivative of (x^2+1/(sqrt(x^2+1)))

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

tangent 3/(4x+1)

t an g e n t \frac{3}{4 x + 1}

derivative of sin^2(2x^2)

\frac{d}{d x} (sin^{2} (2) x^{2})

integral of 5/(1+4x^2)

\int \frac{5}{1 + 4 x ^{2}} d x

integral of sqrt(6x+7)

\int 6 x + 7 d x