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人気のある微分積分 >

integral of (6(1+s))/(s(s^2+7s+6))

解

\int \frac{6 ( 1 + s )}{s ( s ^{2} + 7 s + 6 )} d s

解

ln ∣ s ∣ - ln ∣ s + 6 ∣ + C

解答ステップ

\int \frac{6 ( 1 + s )}{s ( s ^{2} + 7 s + 6 )} d s

定数を除く:

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

= 6 \cdot \int \frac{1 + s}{s ( s ^{2} + 7 s + 6 )} d s

以下の部分分数を得る:

\frac{1 + s}{s ( s ^{2} + 7 s + 6 )} : \frac{1}{6 s} - \frac{1}{6 ( s + 6 )}

= 6 \cdot \int \frac{1}{6 s} - \frac{1}{6 ( s + 6 )} d s

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

= 6 (\int \frac{1}{6 s} d s - \int \frac{1}{6 ( s + 6 )} d s)

\int \frac{1}{6 s} d s = \frac{1}{6} ln ∣ s ∣

\int \frac{1}{6 ( s + 6 )} d s = \frac{1}{6} ln ∣ s + 6 ∣

= 6 (\frac{1}{6} ln ∣ s ∣ - \frac{1}{6} ln ∣ s + 6 ∣)

簡素化

6 (\frac{1}{6} ln ∣ s ∣ - \frac{1}{6} ln ∣ s + 6 ∣) : ln ∣ s ∣ - ln ∣ s + 6 ∣

= ln ∣ s ∣ - ln ∣ s + 6 ∣

定数を解答に追加する

= ln ∣ s ∣ - ln ∣ s + 6 ∣ + C

グラフ

人気の例

integral of sqrt(1-z^2)

\int 1 - z^{2}

limit as h approaches 0 of 1/(sqrt(h))

h \to 0 lim (\frac{1}{h})

limit as x approaches 0 of x/(sin(5x))

x \to 0 lim (\frac{x}{sin ( 5 x )})

integral from 0 to 4 of 1/(x^2-x-2)

\int_{0}^{4} \frac{1}{x ^{2} - x - 2} d x

integral from 0 to pi of x^2cos(nx)

\int_{0}^{π} x^{2} cos (n x) d x