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人気のある微分積分 >

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

解

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

解

\frac{ln ( 5 + 1 ) - ln ( ( 5 - 1 ) ( 2 + 1 ) ) + ln ( 2 - 1 )}{2}

+1

十進法表記

- 0.40016 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{5}^{2} \frac{1}{u ^{2} - 1} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{2}^{5} \frac{1}{u ^{2} - 1} d u

因数

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

= - \int_{2}^{5} \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

= - (- \int_{2}^{5} \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{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{2}^{5})

簡素化

- (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{2}^{5}) : [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{2}^{5}

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

限度を計算する:

\frac{ln ( 5 + 1 ) - ln ( 5 - 1 ) - ln ( 2 + 1 ) + ln ( 2 - 1 )}{2}

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

簡素化

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

グラフ

人気の例

integral from 2 to 3 of 95+768x^2-16x

\int_{2}^{3} 95 + 768 x^{2} - 16 x d x

integral from a to b of sin(x^2)

\int_{a}^{b} sin (x^{2}) d x

integral from 0 to a of xe^{-x}

\int_{0}^{a} x e^{- x} d x

integral from 0 to ln(A) of 6e^{3x}

\int_{0}^{l n (A)} 6 e^{3 x} d x

integral from cos(x) to 1 of sqrt(1-x^2)

\int_{c o s (x)}^{1} 1 - x^{2} d x