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人気のある微分積分 >

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

解

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

解

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

解答ステップ

\int \frac{2 x ^{2}}{( 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

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

u置換積分法を適用する

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

拡張

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

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

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

\int \frac{u ^{2}}{u ^{2} + 4} d u = 4 (- \frac{1}{2} ln (\frac{u + 4 + u ^{2}}{2}) + \frac{1}{8} u 4 + u^{2})

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

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

= 2 (4 (- \frac{1}{2} ln (\frac{u + 4 + u ^{2}}{2}) + \frac{1}{8} u 4 + u^{2}) + 2 u^{2} + 4 + ln (\frac{1}{2} u + 4 + u^{2}))

代用を戻す

u = x - 1

= 2 4 - \frac{1}{2} ln \frac{x - 1 + 4 + ( x - 1 ) ^{2}}{2} + \frac{1}{8} (x - 1) 4 + (x - 1)^{2} + 2 (x - 1)^{2} + 4 + ln (\frac{1}{2} x - 1 + 4 + (x - 1)^{2})

簡素化

2 4 - \frac{1}{2} ln \frac{x - 1 + 4 + ( x - 1 ) ^{2}}{2} + \frac{1}{8} (x - 1) 4 + (x - 1)^{2} + 2 (x - 1)^{2} + 4 + ln (\frac{1}{2} x - 1 + 4 + (x - 1)^{2}) : - 2 ln (\frac{1}{2} x + x^{2} - 2 x + 5 - 1) + x x^{2} - 2 x + 5 + 3 x^{2} - 2 x + 5

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

定数を解答に追加する

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

グラフ

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