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人気のある微分積分 >

integral from 0 to 1 of (12)/(4x^2+5x+1)

解

\int_{0}^{1} \frac{12}{4 x ^{2} + 5 x + 1} d x

解

- 4 ln (\frac{2}{5})

+1

十進法表記

3.66516 \dots

解答ステップ

\int_{0}^{1} \frac{12}{4 x ^{2} + 5 x + 1} d x

定数を除く:

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

= 12 \cdot \int_{0}^{1} \frac{1}{4 x ^{2} + 5 x + 1} d x

平方完成する

4 x^{2} + 5 x + 1 : 4 (x + \frac{5}{8})^{2} - \frac{9}{16}

= 12 \cdot \int_{0}^{1} \frac{1}{4 ( x + \frac{5}{8} ) ^{2} - \frac{9}{16}} d x

u置換積分法を適用する

= 12 \cdot \int_{\frac{5}{8}}^{\frac{13}{8}} \frac{16}{64 u ^{2} - 9} d u

定数を除く:

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

= 12 \cdot 16 \cdot \int_{\frac{5}{8}}^{\frac{13}{8}} \frac{1}{64 u ^{2} - 9} d u

置換積分法を適用する

= 12 \cdot 16 \cdot \int_{\frac{5}{3}}^{\frac{13}{3}} \frac{1}{24 ( v ^{2} - 1 )} d v

定数を除く:

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

= 12 \cdot 16 \cdot \frac{1}{24} \cdot \int_{\frac{5}{3}}^{\frac{13}{3}} \frac{1}{v ^{2} - 1} d v

因数

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

= 12 \cdot 16 \cdot \frac{1}{24} \cdot \int_{\frac{5}{3}}^{\frac{13}{3}} \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

= 12 \cdot 16 \cdot \frac{1}{24} (- \int_{\frac{5}{3}}^{\frac{13}{3}} \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}

= 12 \cdot 16 \cdot \frac{1}{24} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{\frac{13}{3}})

簡素化

12 \cdot 16 \cdot \frac{1}{24} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{\frac{13}{3}}) : - 8 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{5}{3}}^{\frac{13}{3}}

= - 8 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{\frac{5}{3}}^{\frac{13}{3}}

限度を計算する:

\frac{ln ( \frac{16}{3} ) - ln ( \frac{10}{3} ) - ln ( \frac{8}{3} ) + ln ( \frac{2}{3} )}{2}

= - 8 \cdot \frac{ln ( \frac{16}{3} ) - ln ( \frac{10}{3} ) - ln ( \frac{8}{3} ) + ln ( \frac{2}{3} )}{2}

簡素化

= - 4 ln (\frac{2}{5})

グラフ

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