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 (2x^3+x^2-6x+7)/(x^2+x-6)

解

\int \frac{2 x ^{3} + x ^{2} - 6 x + 7}{x ^{2} + x - 6} d x

解

4 (\frac{1}{16} (2 x + 1)^{2} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{8} (7 ln 4 x^{2} + 4 x - 24 + ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + C

解答ステップ

\int \frac{2 x ^{3} + x ^{2} - 6 x + 7}{x ^{2} + x - 6} d x

平方完成する

x^{2} + x - 6 : (x + \frac{1}{2})^{2} - \frac{25}{4}

= \int \frac{2 x ^{3} + x ^{2} - 6 x + 7}{( x + \frac{1}{2} ) ^{2} - \frac{25}{4}} d x

u置換積分法を適用する

= \int \frac{4 ( - 6 u + 2 ( u - \frac{1}{2} ) ^{3} + ( u - \frac{1}{2} ) ^{2} + 10 )}{4 u ^{2} - 25} d u

定数を除く:

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

= 4 \cdot \int \frac{- 6 u + 2 ( u - \frac{1}{2} ) ^{3} + ( u - \frac{1}{2} ) ^{2} + 10}{4 u ^{2} - 25} d u

長除法

\frac{- 6 u + 2 ( u - \frac{1}{2} ) ^{3} + ( u - \frac{1}{2} ) ^{2} + 10}{4 u ^{2} - 25} : \frac{u}{2} - \frac{1}{2} + \frac{14 u - 5}{2 ( 4 u ^{2} - 25 )}

= 4 \cdot \int \frac{u}{2} - \frac{1}{2} + \frac{14 u - 5}{2 ( 4 u ^{2} - 25 )} 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

= 4 (\int \frac{u}{2} d u - \int \frac{1}{2} d u + \int \frac{14 u - 5}{2 ( 4 u ^{2} - 25 )} d u)

\int \frac{u}{2} d u = \frac{u ^{2}}{4}

\int \frac{1}{2} d u = \frac{1}{2} u

\int \frac{14 u - 5}{2 ( 4 u ^{2} - 25 )} d u = \frac{1}{2} (\frac{7}{4} ln 4 u^{2} - 25 + \frac{1}{4} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1))

= 4 (\frac{u ^{2}}{4} - \frac{1}{2} u + \frac{1}{2} (\frac{7}{4} ln 4 u^{2} - 25 + \frac{1}{4} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)))

代用を戻す

u = x + \frac{1}{2}

= 4 (\frac{( x + \frac{1}{2} ) ^{2}}{4} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{2} (\frac{7}{4} ln 4 (x + \frac{1}{2})^{2} - 25 + \frac{1}{4} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1)))

簡素化

4 (\frac{( x + \frac{1}{2} ) ^{2}}{4} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{2} (\frac{7}{4} ln 4 (x + \frac{1}{2})^{2} - 25 + \frac{1}{4} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1))) : 4 (\frac{1}{16} (2 x + 1)^{2} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{8} (7 ln 4 x^{2} + 4 x - 24 + ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

= 4 (\frac{1}{16} (2 x + 1)^{2} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{8} (7 ln 4 x^{2} + 4 x - 24 + ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

定数を解答に追加する

= 4 (\frac{1}{16} (2 x + 1)^{2} - \frac{1}{2} (x + \frac{1}{2}) + \frac{1}{8} (7 ln 4 x^{2} + 4 x - 24 + ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + C

グラフ

人気の例

勾配 (-2.7)(20.16)

s l o p e (- 2.7) (20.16)

limit as x approaches 2-of x^2-1

x \to 2 - lim (x^{2} - 1)

integral of \sqrt[3]{x}-4x

\int 3 x - 4 x d x

limit as x approaches 0 of (1+ax)^{1/x}

x \to 0 lim ((1 + a x)^{\frac{1}{x}})

derivative 3sin(x)ln(x)

d er i v a t i v e 3 sin (x) ln (x)