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人気のある微分積分 >

integral from 2 to 3 of 1/(x^2+3x+2)

解

\int_{2}^{3} \frac{1}{x ^{2} + 3 x + 2} d x

解

- ln (\frac{15}{16})

+1

十進法表記

0.06453 \dots

解答ステップ

\int_{2}^{3} \frac{1}{x ^{2} + 3 x + 2} d x

平方完成する

x^{2} + 3 x + 2 : (x + \frac{3}{2})^{2} - \frac{1}{4}

= \int_{2}^{3} \frac{1}{( x + \frac{3}{2} ) ^{2} - \frac{1}{4}} d x

u置換積分法を適用する

= \int_{\frac{7}{2}}^{\frac{9}{2}} \frac{4}{4 u ^{2} - 1} d u

定数を除く:

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

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

因数

4 u^{2} - 1 : - (- 4 u^{2} + 1)

= 4 \cdot \int_{\frac{7}{2}}^{\frac{9}{2}} \frac{1}{- ( - 4 u ^{2} + 1 )} d u

定数を除く:

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

= 4 (- \int_{\frac{7}{2}}^{\frac{9}{2}} \frac{1}{- 4 u ^{2} + 1} d u)

置換積分法を適用する

= 4 (- \int_{7}^{9} \frac{1}{2 ( - v ^{2} + 1 )} d v)

定数を除く:

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

= 4 (- \frac{1}{2} \cdot \int_{7}^{9} \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}

= 4 (- \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{7}^{9})

簡素化

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

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

限度を計算する:

\frac{ln ( 10 ) - 6 ln ( 2 ) + ln ( 6 )}{2}

= - 2 \cdot \frac{ln ( 10 ) - 6 ln ( 2 ) + ln ( 6 )}{2}

簡素化

= - ln (\frac{15}{16})

グラフ

人気の例

integral from y^2 to y of sqrt(y/x)

\int_{y^{2}}^{y} \frac{y}{x} d x

integral from-1 to 1 of (x^2+x)

\int_{- 1}^{1} (x^{2} + x) d x

integral from 0 to 2 of x^2+5

\int_{0}^{2} x^{2} + 5 d x

integral from 0 to x/2 of cos(x)

\int_{0}^{\frac{x}{2}} cos (x) d x

integral from 0 to pi of (2x-sin(x))

\int_{0}^{π} (2 x - sin (x)) d x