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 ln(1+x^4)

解

\int_{0}^{1} ln (1 + x^{4}) d x

解

2 arctan (1 + 2) + 2 arctan (2 - 1) + \frac{ln ( 3 + 2 2 )}{2} + ln (2) - 4

+1

十進法表記

0.16103 \dots

解答ステップ

\int_{0}^{1} ln (1 + x^{4}) d x

部分積分を適用する

= [x ln (1 + x^{4}) - \int \frac{4 x ^{4}}{1 + x ^{4}} d x]_{0}^{1}

\int \frac{4 x ^{4}}{1 + x ^{4}} d x = - \frac{1}{2} ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1) + \frac{1}{2} ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1) + 4 x

= [x ln (1 + x^{4}) - (- \frac{1}{2} ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1) + \frac{1}{2} ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1) + 4 x)]_{0}^{1}

簡素化

= [x ln (1 + x^{4}) + \frac{1}{2} ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1) - \frac{1}{2} ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1) - 4 x]_{0}^{1}

限度を計算する:

2 arctan (1 + 2) + 2 arctan (2 - 1) + \frac{ln ( 4 + 2 2 ) - ln ( 4 - 2 2 )}{2} + ln (2) - 4

= 2 arctan (1 + 2) + 2 arctan (2 - 1) + \frac{ln ( 4 + 2 2 ) - ln ( 4 - 2 2 )}{2} + ln (2) - 4

簡素化

= 2 arctan (1 + 2) + 2 arctan (2 - 1) + \frac{ln ( 3 + 2 2 )}{2} + ln (2) - 4

グラフ

人気の例

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

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

integral from 8 to 9 of e^{8-x}

\int_{8}^{9} e^{8 - x} d x

integral from 0 to 0 of sqrt(36-x^2)

\int_{0}^{0} 36 - x^{2} d x

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

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

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

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