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 ((x-8))/(x^2-7x+10)

解

\int \frac{( x - 8 )}{x ^{2} - 7 x + 10} d x

解

2 (\frac{1}{4} ln 4 x^{2} - 28 x + 40 + \frac{3}{4} (ln \frac{2 x - 7}{3} + 1 - ln \frac{2 x - 7}{3} - 1)) + C

解答ステップ

\int \frac{x - 8}{x ^{2} - 7 x + 10} d x

u置換積分法を適用する

= \int \frac{u}{u ^{2} + 9 u + 18} d u

平方完成する

u^{2} + 9 u + 18 : (u + \frac{9}{2})^{2} - \frac{9}{4}

= \int \frac{u}{( u + \frac{9}{2} ) ^{2} - \frac{9}{4}} d u

u置換積分法を適用する

= \int \frac{2 ( 2 v - 9 )}{4 v ^{2} - 9} d v

定数を除く:

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

= 2 \cdot \int \frac{2 v - 9}{4 v ^{2} - 9} d v

拡張

\frac{2 v - 9}{4 v ^{2} - 9} : \frac{2 v}{4 v ^{2} - 9} - \frac{9}{4 v ^{2} - 9}

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

= 2 (\int \frac{2 v}{4 v ^{2} - 9} d v - \int \frac{9}{4 v ^{2} - 9} d v)

\int \frac{2 v}{4 v ^{2} - 9} d v = \frac{1}{4} ln 4 v^{2} - 9

\int \frac{9}{4 v ^{2} - 9} d v = - \frac{3}{4} (ln \frac{2}{3} v + 1 - ln \frac{2}{3} v - 1)

= 2 (\frac{1}{4} ln 4 v^{2} - 9 - (- \frac{3}{4} (ln \frac{2}{3} v + 1 - ln \frac{2}{3} v - 1)))

代用を戻す

= 2 (\frac{1}{4} ln 4 (x - 8 + \frac{9}{2})^{2} - 9 - (- \frac{3}{4} (ln \frac{2}{3} (x - 8 + \frac{9}{2}) + 1 - ln \frac{2}{3} (x - 8 + \frac{9}{2}) - 1)))

簡素化

2 (\frac{1}{4} ln 4 (x - 8 + \frac{9}{2})^{2} - 9 - (- \frac{3}{4} (ln \frac{2}{3} (x - 8 + \frac{9}{2}) + 1 - ln \frac{2}{3} (x - 8 + \frac{9}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 40 + \frac{3}{4} (ln \frac{2 x - 7}{3} + 1 - ln \frac{2 x - 7}{3} - 1))

= 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 40 + \frac{3}{4} (ln \frac{2 x - 7}{3} + 1 - ln \frac{2 x - 7}{3} - 1))

定数を解答に追加する

= 2 (\frac{1}{4} ln 4 x^{2} - 28 x + 40 + \frac{3}{4} (ln \frac{2 x - 7}{3} + 1 - ln \frac{2 x - 7}{3} - 1)) + C

グラフ

人気の例

y^{''}+2y^'+y=2e^x

y^{^{''}} + 2 y^{^{'}} + y = 2 e^{x}

(\partial)/(\partial y)(e^{-2x^2-y^2})

\frac{\partial}{\partial y} (e^{- 2 x^{2} - y^{2}})

limit as x approaches-infinity of bx

x \to - \infty lim (b x)

derivative of ln(sec(x+tan(x)))

\frac{d}{d x} (ln (sec (x) + tan (x)))

integral of 1200e^{0.3t}

\int 1200 e^{0.3 t} d t