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人気のある微分積分 >

integral of (-19x^2+50x-25)/(x^2(3x-5))

解

\int \frac{- 19 x ^{2} + 50 x - 25}{x ^{2} ( 3 x - 5 )} d x

解

\frac{1}{3} (- 21 ln (3 ∣ x ∣) - \frac{15}{x} + 2 ln ∣ 3 x - 5 ∣) + C

解答ステップ

\int \frac{- 19 x ^{2} + 50 x - 25}{x ^{2} ( 3 x - 5 )} d x

u置換積分法を適用する

= \int \frac{- 19 u ^{2} - 40 u + 50}{3 u ( u + 5 ) ^{2}} d u

定数を除く:

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

= \frac{1}{3} \cdot \int \frac{- 19 u ^{2} - 40 u + 50}{u ( u + 5 ) ^{2}} d u

拡張

\frac{- 19 u ^{2} - 40 u + 50}{u ( u + 5 ) ^{2}} : - \frac{19 u ^{2}}{u ( u + 5 ) ^{2}} - \frac{40 u}{u ( u + 5 ) ^{2}} + \frac{50}{u ( u + 5 ) ^{2}}

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

= \frac{1}{3} (- \int \frac{19 u ^{2}}{u ( u + 5 ) ^{2}} d u - \int \frac{40 u}{u ( u + 5 ) ^{2}} d u + \int \frac{50}{u ( u + 5 ) ^{2}} d u)

\int \frac{19 u ^{2}}{u ( u + 5 ) ^{2}} d u = 19 (ln ∣ u + 5 ∣ + \frac{5}{u + 5})

\int \frac{40 u}{u ( u + 5 ) ^{2}} d u = - \frac{40}{u + 5}

\int \frac{50}{u ( u + 5 ) ^{2}} d u = 50 (\frac{1}{25} ln ∣ u ∣ - \frac{1}{25} ln ∣ u + 5 ∣ + \frac{1}{5 ( u + 5 )})

= \frac{1}{3} (- 19 (ln ∣ u + 5 ∣ + \frac{5}{u + 5}) - (- \frac{40}{u + 5}) + 50 (\frac{1}{25} ln ∣ u ∣ - \frac{1}{25} ln ∣ u + 5 ∣ + \frac{1}{5 ( u + 5 )}))

代用を戻す

u = 3 x - 5

= \frac{1}{3} (- 19 (ln ∣ 3 x - 5 + 5 ∣ + \frac{5}{3 x - 5 + 5}) - (- \frac{40}{3 x - 5 + 5}) + 50 (\frac{1}{25} ln ∣ 3 x - 5 ∣ - \frac{1}{25} ln ∣ 3 x - 5 + 5 ∣ + \frac{1}{5 ( 3 x - 5 + 5 )}))

簡素化

\frac{1}{3} (- 19 (ln ∣ 3 x - 5 + 5 ∣ + \frac{5}{3 x - 5 + 5}) - (- \frac{40}{3 x - 5 + 5}) + 50 (\frac{1}{25} ln ∣ 3 x - 5 ∣ - \frac{1}{25} ln ∣ 3 x - 5 + 5 ∣ + \frac{1}{5 ( 3 x - 5 + 5 )})) : \frac{1}{3} (- 21 ln (3 ∣ x ∣) - \frac{15}{x} + 2 ln ∣ 3 x - 5 ∣)

= \frac{1}{3} (- 21 ln (3 ∣ x ∣) - \frac{15}{x} + 2 ln ∣ 3 x - 5 ∣)

定数を解答に追加する

= \frac{1}{3} (- 21 ln (3 ∣ x ∣) - \frac{15}{x} + 2 ln ∣ 3 x - 5 ∣) + C

グラフ

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