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人気のある微分積分 >

integral from 0 to 5 of 6/(7x^2+8x+1)

解

\int_{0}^{5} \frac{6}{7 x ^{2} + 8 x + 1} d x

解

ln (6)

+1

十進法表記

1.79175 \dots

解答ステップ

\int_{0}^{5} \frac{6}{7 x ^{2} + 8 x + 1} d x

定数を除く:

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

= 6 \cdot \int_{0}^{5} \frac{1}{7 x ^{2} + 8 x + 1} d x

平方完成する

7 x^{2} + 8 x + 1 : 7 (x + \frac{4}{7})^{2} - \frac{9}{7}

= 6 \cdot \int_{0}^{5} \frac{1}{7 ( x + \frac{4}{7} ) ^{2} - \frac{9}{7}} d x

u置換積分法を適用する

= 6 \cdot \int_{\frac{4}{7}}^{\frac{39}{7}} \frac{7}{49 u ^{2} - 9} d u

定数を除く:

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

= 6 \cdot 7 \cdot \int_{\frac{4}{7}}^{\frac{39}{7}} \frac{1}{49 u ^{2} - 9} d u

置換積分法を適用する

= 6 \cdot 7 \cdot \int_{\frac{4}{3}}^{13} \frac{1}{21 ( v ^{2} - 1 )} d v

定数を除く:

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

= 6 \cdot 7 \cdot \frac{1}{21} \cdot \int_{\frac{4}{3}}^{13} \frac{1}{v ^{2} - 1} d v

因数

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

= 6 \cdot 7 \cdot \frac{1}{21} \cdot \int_{\frac{4}{3}}^{13} \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

= 6 \cdot 7 \cdot \frac{1}{21} (- \int_{\frac{4}{3}}^{13} \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}

= 6 \cdot 7 \cdot \frac{1}{21} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{4}{3}}^{13})

簡素化

6 \cdot 7 \cdot \frac{1}{21} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{4}{3}}^{13}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{4}{3}}^{13}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{4}{3}}^{13}

限度を計算する:

\frac{ln ( 14 ) - ln ( 12 ) - ln ( \frac{7}{3} ) - ln ( 3 )}{2}

= - 2 \cdot \frac{ln ( 14 ) - ln ( 12 ) - ln ( \frac{7}{3} ) - ln ( 3 )}{2}

簡素化

= ln (6)

グラフ

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