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 of sqrt(5+2x^2)

解

\int 5 + 2 x^{2} d x

解

\frac{1}{2} x 5 + 2 x^{2} + \frac{5}{2 2} ln \frac{2}{5} x + \frac{1}{5} (5 + 2 x^{2}) + C

解答ステップ

\int 5 + 2 x^{2} d x

三角関数による置換を適用する

= \int \frac{5}{2} sec^{3} (u) d u

定数を除く:

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

= \frac{5}{2} \cdot \int sec^{3} (u) d u

積分換算を適用する

= \frac{5}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{5}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arctan (\frac{2}{5} x)

= \frac{5}{2} \frac{sec ^{2} ( arctan ( \frac{2}{5} x ) ) sin ( arctan ( \frac{2}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{5} x)) + sec (arctan (\frac{2}{5} x))

簡素化

\frac{5}{2} \frac{sec ^{2} ( arctan ( \frac{2}{5} x ) ) sin ( arctan ( \frac{2}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{5} x)) + sec (arctan (\frac{2}{5} x)) : \frac{1}{2} x 5 + 2 x^{2} + \frac{5}{2 2} ln \frac{2}{5} x + \frac{1}{5} (5 + 2 x^{2})

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

定数を解答に追加する

= \frac{1}{2} x 5 + 2 x^{2} + \frac{5}{2 2} ln \frac{2}{5} x + \frac{1}{5} (5 + 2 x^{2}) + C

グラフ

人気の例

integral of-5xsin(x^2)

\int - 5 x sin (x^{2}) d x

integral of (28)/(1+16x^2)

\int \frac{28}{1 + 16 x ^{2}} d x

integral of (sin(x))/((1-sin(x))^2)

\int \frac{sin ( x )}{( 1 - sin ( x ) ) ^{2}} d x

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

\int \frac{x + 3}{x ^{2} + x + 1} d x

integral of (2x^2+4x-3)/((x-2)^3)

\int \frac{2 x ^{2} + 4 x - 3}{( x - 2 ) ^{3}} d x