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受欢迎的微积分 >

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

解答

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

解答

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

+1

十进制

- 0.40016 \dots

求解步骤

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

使用换元积分法

= \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

因式分解

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

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

提出常数:

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

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

使用积分常见定则:

\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})

化简

- (- [\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}

带入上下限计算后得:

\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}

化简

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

作图

流行的例子

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

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

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

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