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人気のある微分積分 >

integral of (5x^2)/((x-8)^3)

解

\int \frac{5 x ^{2}}{( x - 8 ) ^{3}} d x

解

5 (ln ∣ x - 8 ∣ - \frac{16}{x - 8} - \frac{32}{( x - 8 ) ^{2}}) + C

解答ステップ

\int \frac{5 x ^{2}}{( x - 8 ) ^{3}} 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{x ^{2}}{( x - 8 ) ^{3}} d x

u置換積分法を適用する

= 5 \cdot \int \frac{( u + 8 ) ^{2}}{u ^{3}} d u

拡張

\frac{( u + 8 ) ^{2}}{u ^{3}} : \frac{1}{u} + \frac{16}{u ^{2}} + \frac{64}{u ^{3}}

= 5 \cdot \int \frac{1}{u} + \frac{16}{u ^{2}} + \frac{64}{u ^{3}} 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 (\int \frac{1}{u} d u + \int \frac{16}{u ^{2}} d u + \int \frac{64}{u ^{3}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{16}{u ^{2}} d u = - \frac{16}{u}

\int \frac{64}{u ^{3}} d u = - \frac{32}{u ^{2}}

= 5 (ln ∣ u ∣ - \frac{16}{u} - \frac{32}{u ^{2}})

代用を戻す

u = x - 8

= 5 (ln ∣ x - 8 ∣ - \frac{16}{x - 8} - \frac{32}{( x - 8 ) ^{2}})

定数を解答に追加する

= 5 (ln ∣ x - 8 ∣ - \frac{16}{x - 8} - \frac{32}{( x - 8 ) ^{2}}) + C

グラフ

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