integral from 0 to x of 5/(1+5x^2)

解

\int_{0}^{x} \frac{5}{1 + 5 x ^{2}} d x

5 arctan (5 x)

解答ステップ

\int_{0}^{x} \frac{5}{1 + 5 x ^{2}} d x

定数を除く:

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

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

置換積分法を適用する

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

定数を除く:

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

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

共通積分を使用する:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 5 \cdot \frac{1}{5} [arctan (u)]_{0}^{5 x}

簡素化

5 \cdot \frac{1}{5} [arctan (u)]_{0}^{5 x} : 5 [arctan (u)]_{0}^{5 x}

= 5 [arctan (u)]_{0}^{5 x}

限度を計算する:

arctan (5 x)

= 5 arctan (5 x)