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 1 of (x^2)/(x^3+7)

解

\int_{0}^{1} \frac{x ^{2}}{x ^{3} + 7} d x

解

ln (2) - \frac{1}{3} ln (7)

+1

十進法表記

0.04451 \dots

解答ステップ

\int_{0}^{1} \frac{x ^{2}}{x ^{3} + 7} d x

u置換積分法を適用する

= \int_{7}^{8} \frac{1}{3 u} d u

定数を除く:

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

= \frac{1}{3} \cdot \int_{7}^{8} \frac{1}{u} d u

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{7}^{8}

限度を計算する:

3 ln (2) - ln (7)

= \frac{1}{3} (3 ln (2) - ln (7))

簡素化

= ln (2) - \frac{1}{3} ln (7)

グラフ

人気の例

integral from-1 to 1 of (4x^3+5x^2-x+1)

\int_{- 1}^{1} (4 x^{3} + 5 x^{2} - x + 1) d x

integral from 0 to 3 of x^3sqrt(9-x^2)

\int_{0}^{3} x^{3} 9 - x^{2} d x

integral from-2 to 0 of (x+1)e^{x^2+2x}

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

integral from 11 to infinity of 1/(x^5)

\int_{11}^{\infty} \frac{1}{x ^{5}} d x

integral from 0 to 1 of (2-x^2)^{3/2}

\int_{0}^{1} (2 - x^{2})^{\frac{3}{2}} d x