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人気のある微分積分 >

integral from 0 to 4 of 5/(6x^2+7x+1)

解

\int_{0}^{4} \frac{5}{6 x ^{2} + 7 x + 1} d x

解

ln (5)

+1

十進法表記

1.60943 \dots

解答ステップ

\int_{0}^{4} \frac{5}{6 x ^{2} + 7 x + 1} d x

定数を除く:

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

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

平方完成する

6 x^{2} + 7 x + 1 : 6 (x + \frac{7}{12})^{2} - \frac{25}{24}

= 5 \cdot \int_{0}^{4} \frac{1}{6 ( x + \frac{7}{12} ) ^{2} - \frac{25}{24}} d x

u置換積分法を適用する

= 5 \cdot \int_{\frac{7}{12}}^{\frac{55}{12}} \frac{24}{144 u ^{2} - 25} d u

定数を除く:

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

= 5 \cdot 24 \cdot \int_{\frac{7}{12}}^{\frac{55}{12}} \frac{1}{144 u ^{2} - 25} d u

置換積分法を適用する

= 5 \cdot 24 \cdot \int_{\frac{7}{5}}^{11} \frac{1}{60 ( v ^{2} - 1 )} d v

定数を除く:

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

= 5 \cdot 24 \cdot \frac{1}{60} \cdot \int_{\frac{7}{5}}^{11} \frac{1}{v ^{2} - 1} d v

因数

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

= 5 \cdot 24 \cdot \frac{1}{60} \cdot \int_{\frac{7}{5}}^{11} \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

= 5 \cdot 24 \cdot \frac{1}{60} (- \int_{\frac{7}{5}}^{11} \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}

= 5 \cdot 24 \cdot \frac{1}{60} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{7}{5}}^{11})

簡素化

5 \cdot 24 \cdot \frac{1}{60} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{7}{5}}^{11}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{7}{5}}^{11}

= - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{7}{5}}^{11}

限度を計算する:

\frac{ln ( 12 ) - ln ( 10 ) - ln ( \frac{12}{5} ) + ln ( \frac{2}{5} )}{2}

= - 2 \cdot \frac{ln ( 12 ) - ln ( 10 ) - ln ( \frac{12}{5} ) + ln ( \frac{2}{5} )}{2}

簡素化

= ln (5)

グラフ

人気の例

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