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受欢迎的微积分 >

integral from 0 to x of 1/(t^2-25)

解答

\int_{0}^{x} \frac{1}{t ^{2} - 25} d t

解答

- \frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{10}

求解步骤

\int_{0}^{x} \frac{1}{t ^{2} - 25} d t

使用换元积分法

= \int_{0}^{\frac{x}{5}} \frac{1}{5 ( u ^{2} - 1 )} d u

提出常数:

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

= \frac{1}{5} \cdot \int_{0}^{\frac{x}{5}} \frac{1}{u ^{2} - 1} d u

因式分解

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

= \frac{1}{5} \cdot \int_{0}^{\frac{x}{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

= \frac{1}{5} (- \int_{0}^{\frac{x}{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{1}{5} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}})

化简

\frac{1}{5} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}}) : - \frac{1}{5} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}}

= - \frac{1}{5} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}}

带入上下限计算后得:

\frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{2}

= - \frac{1}{5} \cdot \frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{2}

化简

= - \frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{10}

流行的例子

y^'=y+3te^t,y(0)=4

y^{^{'}} = y + 3 t e^{t}, y (0) = 4

limit as x approaches 3 of sqrt(x-8)

x \to 3 lim (x - 8)

integral of (s^3+3)^2(1-s^2)^3

\int (s^{3} + 3)^{2} (1 - s^{2})^{3} d s

y^{''}-16y=2e^{4x}

y^{^{''}} - 16 y = 2 e^{4 x}

integral from 1 to 2 of sqrt(t)ln(t)

\int_{1}^{2} t ln (t) d t