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

解

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

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

解答ステップ

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

置換積分法を適用する

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

因数

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

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

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

簡素化

\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}}

限度を計算する:

\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}

簡素化

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