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人気のある微分積分 >

integral of (-3x)/(sqrt((x-1)^2+4))

解

\int \frac{- 3 x}{( x - 1 ) ^{2} + 4} d x

解

- 3 (x^{2} - 2 x + 5 + ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5)) + C

解答ステップ

\int \frac{- 3 x}{( x - 1 ) ^{2} + 4} d x

定数を除く:

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

= - 3 \cdot \int \frac{x}{( x - 1 ) ^{2} + 4} d x

拡張

(x - 1)^{2} + 4 : x^{2} - 2 x + 5

= - 3 \cdot \int \frac{x}{x ^{2} - 2 x + 5} d x

平方完成する

x^{2} - 2 x + 5 : (x - 1)^{2} + 4

= - 3 \cdot \int \frac{x}{( x - 1 ) ^{2} + 4} d x

u置換積分法を適用する

= - 3 \cdot \int \frac{u + 1}{u ^{2} + 4} d u

拡張

\frac{u + 1}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} + \frac{1}{u ^{2} + 4}

= - 3 \cdot \int \frac{u}{u ^{2} + 4} + \frac{1}{u ^{2} + 4} 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

= - 3 (\int \frac{u}{u ^{2} + 4} d u + \int \frac{1}{u ^{2} + 4} d u)

\int \frac{u}{u ^{2} + 4} d u = u^{2} + 4

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

= - 3 (u^{2} + 4 + ln (\frac{1}{2} u + 4 + u^{2}))

代用を戻す

u = x - 1

= - 3 ((x - 1)^{2} + 4 + ln (\frac{1}{2} x - 1 + 4 + (x - 1)^{2}))

簡素化

- 3 ((x - 1)^{2} + 4 + ln (\frac{1}{2} x - 1 + 4 + (x - 1)^{2})) : - 3 (x^{2} - 2 x + 5 + ln (\frac{1}{2} x - 1 + x^{2} - 2 x + 5))

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

定数を解答に追加する

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

グラフ

人気の例

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\frac{d y}{d t} = 3 y + y^{4}, y (0) = 1

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