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人気のある微分積分 >

integral of (2x)/(sqrt(9x^2-6x-8))

解

\int \frac{2 x}{9 x ^{2} - 6 x - 8} d x

解

\frac{2}{9} 3 (x - \frac{1}{3})^{2} - 1 + ln (x - \frac{1}{3})^{2} - 1 + x - \frac{1}{3} + C

解答ステップ

\int \frac{2 x}{9 x ^{2} - 6 x - 8} 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}{9 x ^{2} - 6 x - 8} d x

平方完成する

9 x^{2} - 6 x - 8 : 9 (x - \frac{1}{3})^{2} - 9

= 2 \cdot \int \frac{x}{9 ( x - \frac{1}{3} ) ^{2} - 9} d x

u置換積分法を適用する

= 2 \cdot \int \frac{3 u + 1}{9 u ^{2} - 1} d u

定数を除く:

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

= 2 \cdot \frac{1}{9} \cdot \int \frac{3 u + 1}{u ^{2} - 1} d u

拡張

\frac{3 u + 1}{u ^{2} - 1} : \frac{3 u}{u ^{2} - 1} + \frac{1}{u ^{2} - 1}

= 2 \cdot \frac{1}{9} \cdot \int \frac{3 u}{u ^{2} - 1} + \frac{1}{u ^{2} - 1} 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 \cdot \frac{1}{9} (\int \frac{3 u}{u ^{2} - 1} d u + \int \frac{1}{u ^{2} - 1} d u)

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

\int \frac{1}{u ^{2} - 1} d u = ln u^{2} - 1 + u

= 2 \cdot \frac{1}{9} (3 u^{2} - 1 + ln u^{2} - 1 + u)

代用を戻す

u = x - \frac{1}{3}

= 2 \cdot \frac{1}{9} 3 (x - \frac{1}{3})^{2} - 1 + ln (x - \frac{1}{3})^{2} - 1 + x - \frac{1}{3}

簡素化

2 \cdot \frac{1}{9} 3 (x - \frac{1}{3})^{2} - 1 + ln (x - \frac{1}{3})^{2} - 1 + x - \frac{1}{3} : \frac{2}{9} 3 (x - \frac{1}{3})^{2} - 1 + ln (x - \frac{1}{3})^{2} - 1 + x - \frac{1}{3}

= \frac{2}{9} 3 (x - \frac{1}{3})^{2} - 1 + ln (x - \frac{1}{3})^{2} - 1 + x - \frac{1}{3}

定数を解答に追加する

= \frac{2}{9} 3 (x - \frac{1}{3})^{2} - 1 + ln (x - \frac{1}{3})^{2} - 1 + x - \frac{1}{3} + C

グラフ

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