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人気のある微分積分 >

integral of 5/(x(x^2+3)^2)

解

\int \frac{5}{x ( x ^{2} + 3 ) ^{2}} d x

解

\frac{5}{9} ln (x) - \frac{5}{18} ln x^{2} + 3 + \frac{5}{6 ( x ^{2} + 3 )} + C

解答ステップ

\int \frac{5}{x ( x ^{2} + 3 ) ^{2}} 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 ^{2} + 3 ) ^{2}} d x

u置換積分法を適用する

= 5 \cdot \int \frac{1}{2 ( u - 3 ) u ^{2}} 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}{2} \cdot \int \frac{1}{( u - 3 ) u ^{2}} d u

以下の部分分数を得る:

\frac{1}{( u - 3 ) u ^{2}} : \frac{1}{9 ( u - 3 )} - \frac{1}{9 u} - \frac{1}{3 u ^{2}}

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{9 ( u - 3 )} - \frac{1}{9 u} - \frac{1}{3 u ^{2}} 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}{2} (\int \frac{1}{9 ( u - 3 )} d u - \int \frac{1}{9 u} d u - \int \frac{1}{3 u ^{2}} d u)

\int \frac{1}{9 ( u - 3 )} d u = \frac{1}{9} ln ∣ u - 3 ∣

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{3 u ^{2}} d u = - \frac{1}{3 u}

= 5 \cdot \frac{1}{2} (\frac{1}{9} ln ∣ u - 3 ∣ - \frac{1}{9} ln ∣ u ∣ - (- \frac{1}{3 u}))

代用を戻す

u = x^{2} + 3

= 5 \cdot \frac{1}{2} (\frac{1}{9} ln x^{2} + 3 - 3 - \frac{1}{9} ln x^{2} + 3 - (- \frac{1}{3 ( x ^{2} + 3 )}))

簡素化

5 \cdot \frac{1}{2} (\frac{1}{9} ln x^{2} + 3 - 3 - \frac{1}{9} ln x^{2} + 3 - (- \frac{1}{3 ( x ^{2} + 3 )})) : \frac{5}{9} ln (x) - \frac{5}{18} ln x^{2} + 3 + \frac{5}{6 ( x ^{2} + 3 )}

= \frac{5}{9} ln (x) - \frac{5}{18} ln x^{2} + 3 + \frac{5}{6 ( x ^{2} + 3 )}

定数を解答に追加する

= \frac{5}{9} ln (x) - \frac{5}{18} ln x^{2} + 3 + \frac{5}{6 ( x ^{2} + 3 )} + C

グラフ

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