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人気のある微分積分 >

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

解

\int \frac{x}{( x - 1 ) ( x ^{2} + 2 x + 2 ) ^{2}} d x

解

\frac{1}{4} (2 arctan (x + 1) + sin (2 arctan (x + 1))) - \frac{1}{20} sin (2 arctan (x + 1)) (x - 1) - \frac{7}{25} arctan (x + 1) - \frac{1}{5} sin (2 arctan (x + 1)) + \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{50} ln x^{2} + 2 x + 2 + C

解答ステップ

\int \frac{x}{( x - 1 ) ( x ^{2} + 2 x + 2 ) ^{2}} d x

u置換積分法を適用する

= \int \frac{u + 1}{u ( u ^{2} + 4 u + 5 ) ^{2}} d u

拡張

\frac{u + 1}{u ( u ^{2} + 4 u + 5 ) ^{2}} : \frac{u}{u ( u ^{2} + 4 u + 5 ) ^{2}} + \frac{1}{u ( u ^{2} + 4 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

= \int \frac{u}{u ( u ^{2} + 4 u + 5 ) ^{2}} d u + \int \frac{1}{u ( u ^{2} + 4 u + 5 ) ^{2}} d u

\int \frac{u}{u ( u ^{2} + 4 u + 5 ) ^{2}} d u = \frac{1}{2} (arctan (u + 2) + \frac{1}{2} sin (2 arctan (u + 2)))

\int \frac{1}{u ( u ^{2} + 4 u + 5 ) ^{2}} d u = - \frac{1}{20} u sin (2 arctan (u + 2)) - \frac{7}{25} arctan (u + 2) - \frac{1}{5} sin (2 arctan (u + 2)) + \frac{1}{25} ln ∣ u ∣ - \frac{1}{50} ln u^{2} + 4 u + 5

= \frac{1}{2} (arctan (u + 2) + \frac{1}{2} sin (2 arctan (u + 2))) - \frac{1}{20} u sin (2 arctan (u + 2)) - \frac{7}{25} arctan (u + 2) - \frac{1}{5} sin (2 arctan (u + 2)) + \frac{1}{25} ln ∣ u ∣ - \frac{1}{50} ln u^{2} + 4 u + 5

代用を戻す

u = x - 1

= \frac{1}{2} (arctan (x - 1 + 2) + \frac{1}{2} sin (2 arctan (x - 1 + 2))) - \frac{1}{20} (x - 1) sin (2 arctan (x - 1 + 2)) - \frac{7}{25} arctan (x - 1 + 2) - \frac{1}{5} sin (2 arctan (x - 1 + 2)) + \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{50} ln (x - 1)^{2} + 4 (x - 1) + 5

簡素化

\frac{1}{2} (arctan (x - 1 + 2) + \frac{1}{2} sin (2 arctan (x - 1 + 2))) - \frac{1}{20} (x - 1) sin (2 arctan (x - 1 + 2)) - \frac{7}{25} arctan (x - 1 + 2) - \frac{1}{5} sin (2 arctan (x - 1 + 2)) + \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{50} ln (x - 1)^{2} + 4 (x - 1) + 5 : \frac{1}{4} (2 arctan (x + 1) + sin (2 arctan (x + 1))) - \frac{1}{20} sin (2 arctan (x + 1)) (x - 1) - \frac{7}{25} arctan (x + 1) - \frac{1}{5} sin (2 arctan (x + 1)) + \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{50} ln x^{2} + 2 x + 2

= \frac{1}{4} (2 arctan (x + 1) + sin (2 arctan (x + 1))) - \frac{1}{20} sin (2 arctan (x + 1)) (x - 1) - \frac{7}{25} arctan (x + 1) - \frac{1}{5} sin (2 arctan (x + 1)) + \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{50} ln x^{2} + 2 x + 2

定数を解答に追加する

= \frac{1}{4} (2 arctan (x + 1) + sin (2 arctan (x + 1))) - \frac{1}{20} sin (2 arctan (x + 1)) (x - 1) - \frac{7}{25} arctan (x + 1) - \frac{1}{5} sin (2 arctan (x + 1)) + \frac{1}{25} ln ∣ x - 1 ∣ - \frac{1}{50} ln x^{2} + 2 x + 2 + C

グラフ

人気の例

limit as x approaches 0+of 2/x

x \to 0 + lim (\frac{2}{x})

x(dy)/(dx)+8y=e^{x^8}

x \frac{d y}{d x} + 8 y = e^{x^{8}}

limit as x approaches 0+of |1+1/(x-1)|

x \to 0 + lim (1 + \frac{1}{x - 1})

derivative of 2x^2+8x+10

\frac{d}{d x} (2 x^{2} + 8 x + 10)

limit as x approaches-3 of sqrt(7+x)

x \to - 3 lim (7 + x)