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人気のある微分積分 >

integral of 5/(x(x^4+7))

解

\int \frac{5}{x ( x ^{4} + 7 )} d x

解

\frac{5}{7} ln (x) - \frac{5}{28} ln x^{4} + 7 + C

解答ステップ

\int \frac{5}{x ( x ^{4} + 7 )} d x

定数を除く:

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

= 5 \cdot \int \frac{1}{x ( x ^{4} + 7 )} d x

u置換積分法を適用する

= 5 \cdot \int \frac{1}{4 u ( u + 7 )} d u

定数を除く:

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

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 7 )} d u

以下の部分分数を得る:

\frac{1}{u ( u + 7 )} : \frac{1}{7 u} - \frac{1}{7 ( u + 7 )}

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{7 u} - \frac{1}{7 ( u + 7 )} 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

= 5 \cdot \frac{1}{4} (\int \frac{1}{7 u} d u - \int \frac{1}{7 ( u + 7 )} d u)

\int \frac{1}{7 u} d u = \frac{1}{7} ln ∣ u ∣

\int \frac{1}{7 ( u + 7 )} d u = \frac{1}{7} ln ∣ u + 7 ∣

= 5 \cdot \frac{1}{4} (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

代用を戻す

u = x^{4}

= 5 \cdot \frac{1}{4} (\frac{1}{7} ln x^{4} - \frac{1}{7} ln x^{4} + 7)

簡素化

5 \cdot \frac{1}{4} (\frac{1}{7} ln x^{4} - \frac{1}{7} ln x^{4} + 7) : \frac{5}{7} ln (x) - \frac{5}{28} ln x^{4} + 7

= \frac{5}{7} ln (x) - \frac{5}{28} ln x^{4} + 7

定数を解答に追加する

= \frac{5}{7} ln (x) - \frac{5}{28} ln x^{4} + 7 + C

グラフ

人気の例

derivative of (2+e^x/(2-e^{-x)})

\frac{d}{d x} (\frac{2 + e ^{x}}{2 - e ^{- x}})

integral of (x-2)/(x+1)

\int \frac{x - 2}{x + 1} d x

limit as h approaches 0 of (f(h)(8+h)-f(8))/h

h \to 0 lim (\frac{f ( h ) ( 8 + h ) - f ( 8 )}{h})

勾配 (-5,-10),(-1,5)

s l o p e (- 5, - 10), (- 1, 5)

inverselaplace ((2-5s))/(s^2+9)

in v erse l a pl a ce \frac{( 2 - 5 s )}{s ^{2} + 9}