integral from-5 to x of 1/(t^2+4)

解

\int_{- 5}^{x} \frac{1}{t ^{2} + 4} d t

\frac{1}{2} (arctan (\frac{x}{2}) + arctan (\frac{5}{2}))

解答ステップ

\int_{- 5}^{x} \frac{1}{t ^{2} + 4} d t

置換積分法を適用する

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

共通積分を使用する:

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

= \frac{1}{2} [arctan (u)]_{- \frac{5}{2}}^{\frac{x}{2}}

限度を計算する:

arctan (\frac{x}{2}) + arctan (\frac{5}{2})

= \frac{1}{2} (arctan (\frac{x}{2}) + arctan (\frac{5}{2}))