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人気のある微分積分 >

integral of x^2sqrt(x^2-9)

解

\int x^{2} x^{2} - 9 d x

解

81 (\frac{1}{243} x^{3} x^{2} - 9 - \frac{1}{972} x^{3} x^{2} - 9 - \frac{1}{72} x x^{2} - 9 - \frac{1}{8} ln (\frac{1}{3} x^{2} - 9 + x)) + C

解答ステップ

\int x^{2} x^{2} - 9 d x

x^{2} - 9

の共役で乗じる

\frac{x ^{4} - 81}{x ^{2} + 9}

= \int x^{2} \frac{x ^{4} - 81}{x ^{2} + 9} d x

置換積分法を適用する

= \int 81 u^{2} (u + 1) (u - 1) d u

定数を除く:

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

= 81 \cdot \int u^{2} (u + 1) (u - 1) d u

u + 1

の共役で乗じる

\frac{u ^{2} - 1}{u - 1}

= 81 \cdot \int u^{2} \frac{u ^{2} - 1}{u - 1} (u - 1) d u

乗じる

\frac{u ^{2} - 1}{u - 1} (u - 1) : u^{2} - 1

= 81 \cdot \int u^{2} u^{2} - 1 d u

部分積分を適用する

= 81 (\frac{1}{3} u^{3} u^{2} - 1 - \int \frac{u ^{4}}{3 u ^{2} - 1} d u)

\int \frac{u ^{4}}{3 u ^{2} - 1} d u = \frac{1}{12} u^{3} u^{2} - 1 + \frac{1}{8} (u u^{2} - 1 + ln u^{2} - 1 + u)

= 81 (\frac{1}{3} u^{3} u^{2} - 1 - (\frac{1}{12} u^{3} u^{2} - 1 + \frac{1}{8} (u u^{2} - 1 + ln u^{2} - 1 + u)))

代用を戻す

u = \frac{x}{3}

= 81 (\frac{1}{3} (\frac{x}{3})^{3} (\frac{x}{3})^{2} - 1 - (\frac{1}{12} (\frac{x}{3})^{3} (\frac{x}{3})^{2} - 1 + \frac{1}{8} (\frac{x}{3} (\frac{x}{3})^{2} - 1 + ln (\frac{x}{3})^{2} - 1 + \frac{x}{3})))

簡素化

81 (\frac{1}{3} (\frac{x}{3})^{3} (\frac{x}{3})^{2} - 1 - (\frac{1}{12} (\frac{x}{3})^{3} (\frac{x}{3})^{2} - 1 + \frac{1}{8} (\frac{x}{3} (\frac{x}{3})^{2} - 1 + ln (\frac{x}{3})^{2} - 1 + \frac{x}{3}))) : 81 (\frac{1}{243} x^{3} x^{2} - 9 - \frac{1}{972} x^{3} x^{2} - 9 - \frac{1}{72} x x^{2} - 9 - \frac{1}{8} ln (\frac{1}{3} x^{2} - 9 + x))

= 81 (\frac{1}{243} x^{3} x^{2} - 9 - \frac{1}{972} x^{3} x^{2} - 9 - \frac{1}{72} x x^{2} - 9 - \frac{1}{8} ln (\frac{1}{3} x^{2} - 9 + x))

定数を解答に追加する

= 81 (\frac{1}{243} x^{3} x^{2} - 9 - \frac{1}{972} x^{3} x^{2} - 9 - \frac{1}{72} x x^{2} - 9 - \frac{1}{8} ln (\frac{1}{3} x^{2} - 9 + x)) + C

グラフ

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