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人気のある微分積分 >

integral from 0 to 7 of 8/(9x^2+10x+1)

解

\int_{0}^{7} \frac{8}{9 x ^{2} + 10 x + 1} d x

解

3 ln (2)

+1

十進法表記

2.07944 \dots

解答ステップ

\int_{0}^{7} \frac{8}{9 x ^{2} + 10 x + 1} d x

定数を除く:

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

= 8 \cdot \int_{0}^{7} \frac{1}{9 x ^{2} + 10 x + 1} d x

平方完成する

9 x^{2} + 10 x + 1 : 9 (x + \frac{5}{9})^{2} - \frac{16}{9}

= 8 \cdot \int_{0}^{7} \frac{1}{9 ( x + \frac{5}{9} ) ^{2} - \frac{16}{9}} d x

u置換積分法を適用する

= 8 \cdot \int_{\frac{5}{9}}^{\frac{68}{9}} \frac{9}{81 u ^{2} - 16} d u

定数を除く:

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

= 8 \cdot 9 \cdot \int_{\frac{5}{9}}^{\frac{68}{9}} \frac{1}{81 u ^{2} - 16} d u

置換積分法を適用する

= 8 \cdot 9 \cdot \int_{\frac{5}{4}}^{17} \frac{1}{36 ( v ^{2} - 1 )} d v

定数を除く:

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

= 8 \cdot 9 \cdot \frac{1}{36} \cdot \int_{\frac{5}{4}}^{17} \frac{1}{v ^{2} - 1} d v

因数

v^{2} - 1 : - (- v^{2} + 1)

= 8 \cdot 9 \cdot \frac{1}{36} \cdot \int_{\frac{5}{4}}^{17} \frac{1}{- ( - v ^{2} + 1 )} d v

定数を除く:

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

= 8 \cdot 9 \cdot \frac{1}{36} (- \int_{\frac{5}{4}}^{17} \frac{1}{- v ^{2} + 1} d v)

共通積分を使用する:

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

= 8 \cdot 9 \cdot \frac{1}{36} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{5}{4}}^{17})

簡素化

8 \cdot 9 \cdot \frac{1}{36} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{5}{4}}^{17}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{5}{4}}^{17}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{5}{4}}^{17}

限度を計算する:

\frac{- 6 ln ( 2 ) + ln ( 18 ) - ln ( \frac{9}{4} )}{2}

= - 2 \cdot \frac{- 6 ln ( 2 ) + ln ( 18 ) - ln ( \frac{9}{4} )}{2}

簡素化

= 3 ln (2)

グラフ

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