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 1 to 2 of sqrt(x^2+1)

解

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

解

5 + \frac{- 2 - ln ( 1 + 2 ) + ln ( 2 + 5 )}{2}

+1

十進法表記

1.81009 \dots

解答ステップ

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

三角関数による置換を適用する

= \int_{\frac{π}{4}}^{a r c t a n (2)} sec^{3} (u) d u

積分換算を適用する

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (2)} + \frac{1}{2} \cdot \int_{\frac{π}{4}}^{a r c t a n (2)} sec (u) d u

\int_{\frac{π}{4}}^{a r c t a n (2)} sec (u) d u = ln (2 + 5) - ln (1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (1 + 2))

簡素化

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (1 + 2)) : [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (1 + 2))

= [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (1 + 2))

限度を計算する:

5 - \frac{1}{2}

= 5 - \frac{1}{2} + \frac{1}{2} (ln (2 + 5) - ln (1 + 2))

簡素化

= 5 + \frac{- 2 - ln ( 1 + 2 ) + ln ( 2 + 5 )}{2}

グラフ

人気の例

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

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

integral from-1 to 1 of t^7(1+t^8)^7

\int_{- 1}^{1} t^{7} (1 + t^{8})^{7} d t

integral from 0 to 1 of xarctan(x^2)

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

integral from 0 to 1 of-x^2+2x

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

integral from 0 to ln(2x) of sin(6e^t)

\int_{0}^{l n (2 x)} sin (6 e^{t}) d t