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人気のある微分積分 >

integral from 0 to 6 of 7/(8x^2+9x+1)

解

\int_{0}^{6} \frac{7}{8 x ^{2} + 9 x + 1} d x

解

ln (7)

+1

十進法表記

1.94591 \dots

解答ステップ

\int_{0}^{6} \frac{7}{8 x ^{2} + 9 x + 1} d x

定数を除く:

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

= 7 \cdot \int_{0}^{6} \frac{1}{8 x ^{2} + 9 x + 1} d x

平方完成する

8 x^{2} + 9 x + 1 : 8 (x + \frac{9}{16})^{2} - \frac{49}{32}

= 7 \cdot \int_{0}^{6} \frac{1}{8 ( x + \frac{9}{16} ) ^{2} - \frac{49}{32}} d x

u置換積分法を適用する

= 7 \cdot \int_{\frac{9}{16}}^{\frac{105}{16}} \frac{32}{256 u ^{2} - 49} d u

定数を除く:

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

= 7 \cdot 32 \cdot \int_{\frac{9}{16}}^{\frac{105}{16}} \frac{1}{256 u ^{2} - 49} d u

置換積分法を適用する

= 7 \cdot 32 \cdot \int_{\frac{9}{7}}^{15} \frac{1}{112 ( v ^{2} - 1 )} d v

定数を除く:

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

= 7 \cdot 32 \cdot \frac{1}{112} \cdot \int_{\frac{9}{7}}^{15} \frac{1}{v ^{2} - 1} d v

因数

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

= 7 \cdot 32 \cdot \frac{1}{112} \cdot \int_{\frac{9}{7}}^{15} \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

= 7 \cdot 32 \cdot \frac{1}{112} (- \int_{\frac{9}{7}}^{15} \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}

= 7 \cdot 32 \cdot \frac{1}{112} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{9}{7}}^{15})

簡素化

7 \cdot 32 \cdot \frac{1}{112} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{9}{7}}^{15}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{9}{7}}^{15}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{9}{7}}^{15}

限度を計算する:

\frac{4 ln ( 2 ) - ln ( 14 ) - ln ( \frac{16}{7} ) + ln ( \frac{2}{7} )}{2}

= - 2 \cdot \frac{4 ln ( 2 ) - ln ( 14 ) - ln ( \frac{16}{7} ) + ln ( \frac{2}{7} )}{2}

簡素化

= ln (7)

グラフ

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