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(x))/(\sqrt[3]{x)+1}

解

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

解

6 (\frac{1}{7} x^{\frac{7}{6}} - \frac{1}{5} x^{\frac{5}{6}} + \frac{1}{3} x - 6 x + arctan (6 x)) + C

解答ステップ

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

u置換積分法を適用する

= \int \frac{6 u ^{8}}{u ^{2} + 1} d u

定数を除く:

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

= 6 \cdot \int \frac{u ^{8}}{u ^{2} + 1} d u

長除法

\frac{u ^{8}}{u ^{2} + 1} : u^{6} - u^{4} + u^{2} - 1 + \frac{1}{u ^{2} + 1}

= 6 \cdot \int u^{6} - u^{4} + u^{2} - 1 + \frac{1}{u ^{2} + 1} d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int u^{6} d u - \int u^{4} d u + \int u^{2} d u - \int 1 d u + \int \frac{1}{u ^{2} + 1} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

\int u^{4} d u = \frac{u ^{5}}{5}

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 6 (\frac{u ^{7}}{7} - \frac{u ^{5}}{5} + \frac{u ^{3}}{3} - u + arctan (u))

代用を戻す

u = 6 x

= 6 (\frac{( 6 x ) ^{7}}{7} - \frac{( 6 x ) ^{5}}{5} + \frac{( 6 x ) ^{3}}{3} - 6 x + arctan (6 x))

簡素化

6 (\frac{( 6 x ) ^{7}}{7} - \frac{( 6 x ) ^{5}}{5} + \frac{( 6 x ) ^{3}}{3} - 6 x + arctan (6 x)) : 6 (\frac{1}{7} x^{\frac{7}{6}} - \frac{1}{5} x^{\frac{5}{6}} + \frac{1}{3} x - 6 x + arctan (6 x))

= 6 (\frac{1}{7} x^{\frac{7}{6}} - \frac{1}{5} x^{\frac{5}{6}} + \frac{1}{3} x - 6 x + arctan (6 x))

定数を解答に追加する

= 6 (\frac{1}{7} x^{\frac{7}{6}} - \frac{1}{5} x^{\frac{5}{6}} + \frac{1}{3} x - 6 x + arctan (6 x)) + C

グラフ

人気の例

limit as x approaches 2 of 4x^3

x \to 2 lim (4 x^{3})

derivative arccos(-3x)

d er i v a t i v e arccos (- 3 x)

derivative of 6.687(0.931^x)

\frac{d}{d x} (6.687 (0.931)^{x})

sum from n=0 to infinity of (x+5)^n

n = 0 \sum \infty (x + 5)^{n}

derivative of (2x/(sqrt(x^2+1)))

\frac{d}{d x} (\frac{2 x}{x ^{2} + 1})