ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

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

解

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

解

\frac{π}{4 2}

+1

十進法表記

0.55536 \dots

解答ステップ

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

置換積分法を適用する

= \int_{0}^{1} \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_{0}^{1} \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)]_{0}^{1}

限度を計算する:

\frac{π}{4}

= \frac{1}{2} \cdot \frac{π}{4}

簡素化

= \frac{π}{4 2}

グラフ

人気の例

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

\int_{0}^{1} (x + y)^{2} d x

integral from 0 to z of z

\int_{0}^{z} z d z

integral from 1 to x of 2-(2x-2ln(x))/x

\int_{1}^{x} 2 - \frac{2 x - 2 ln ( x )}{x} d x

integral from 1 to 3 of 3/(x^3)

\int_{1}^{3} \frac{3}{x ^{3}} d x

integral from 1 to 2 of x(x^2-1)^7

\int_{1}^{2} x (x^{2} - 1)^{7} d x