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人気のある微分積分 >

integral from-1 to 0 of 1/(4x^2-9)

解

\int_{- 1}^{0} \frac{1}{4 x ^{2} - 9} d x

解

- \frac{ln ( 5 )}{12}

+1

十進法表記

- 0.13411 \dots

解答ステップ

\int_{- 1}^{0} \frac{1}{4 x ^{2} - 9} d x

置換積分法を適用する

= \int_{- \frac{2}{3}}^{0} \frac{1}{6 ( u ^{2} - 1 )} d u

定数を除く:

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

= \frac{1}{6} \cdot \int_{- \frac{2}{3}}^{0} \frac{1}{u ^{2} - 1} d u

因数

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

= \frac{1}{6} \cdot \int_{- \frac{2}{3}}^{0} \frac{1}{- ( - u ^{2} + 1 )} d u

定数を除く:

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

= \frac{1}{6} (- \int_{- \frac{2}{3}}^{0} \frac{1}{- u ^{2} + 1} d u)

共通積分を使用する:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{6} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{0})

簡素化

\frac{1}{6} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{0}) : - \frac{1}{6} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{0}

= - \frac{1}{6} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{0}

限度を計算する:

- \frac{- ln ( 3 ) - ln ( \frac{5}{3} )}{2}

= - \frac{1}{6} (- \frac{- ln ( 3 ) - ln ( \frac{5}{3} )}{2})

簡素化

= - \frac{ln ( 5 )}{12}

グラフ

人気の例

integral from 1 to e^x of 4ln(t)

\int_{1}^{e^{x}} 4 ln (t) d t

integral from-3 to 5 of (4x-3)^3

\int_{- 3}^{5} (4 x - 3)^{3} d x

integral from 0 to pi/2 of 1/(cos(x))

\int_{0}^{\frac{π}{2}} \frac{1}{cos ( x )} d x

integral from-2 to-1 of 1/((11+5x)^3)

\int_{- 2}^{- 1} \frac{1}{( 11 + 5 x ) ^{3}} d x

integral from-1 to 5 of (x^2-4x-5)

\int_{- 1}^{5} (x^{2} - 4 x - 5) d x