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-17)/(x^2+x-12)

解

\int \frac{x - 17}{x ^{2} + x - 12} d x

解

2 (\frac{1}{4} ln 4 x^{2} + 4 x - 48 + \frac{5}{4} (ln \frac{8}{7} + \frac{2}{7} x - ln - \frac{6}{7} + \frac{2}{7} x)) + C

解答ステップ

\int \frac{x - 17}{x ^{2} + x - 12} d x

平方完成する

x^{2} + x - 12 : (x + \frac{1}{2})^{2} - \frac{49}{4}

= \int \frac{x - 17}{( x + \frac{1}{2} ) ^{2} - \frac{49}{4}} d x

u置換積分法を適用する

= \int \frac{2 ( 2 u - 35 )}{4 u ^{2} - 49} d u

定数を除く:

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

= 2 \cdot \int \frac{2 u - 35}{4 u ^{2} - 49} d u

拡張

\frac{2 u - 35}{4 u ^{2} - 49} : \frac{2 u}{4 u ^{2} - 49} - \frac{35}{4 u ^{2} - 49}

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 u}{4 u ^{2} - 49} d u - \int \frac{35}{4 u ^{2} - 49} d u)

\int \frac{2 u}{4 u ^{2} - 49} d u = \frac{1}{4} ln 4 u^{2} - 49

\int \frac{35}{4 u ^{2} - 49} d u = - \frac{5}{4} (ln \frac{2}{7} u + 1 - ln \frac{2}{7} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 49 - (- \frac{5}{4} (ln \frac{2}{7} u + 1 - ln \frac{2}{7} u - 1)))

代用を戻す

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

= 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 49 - (- \frac{5}{4} (ln \frac{2}{7} (x + \frac{1}{2}) + 1 - ln \frac{2}{7} (x + \frac{1}{2}) - 1)))

簡素化

2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 49 - (- \frac{5}{4} (ln \frac{2}{7} (x + \frac{1}{2}) + 1 - ln \frac{2}{7} (x + \frac{1}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 48 + \frac{5}{4} (ln \frac{8}{7} + \frac{2}{7} x - ln - \frac{6}{7} + \frac{2}{7} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 48 + \frac{5}{4} (ln \frac{8}{7} + \frac{2}{7} x - ln - \frac{6}{7} + \frac{2}{7} x))

定数を解答に追加する

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 48 + \frac{5}{4} (ln \frac{8}{7} + \frac{2}{7} x - ln - \frac{6}{7} + \frac{2}{7} x)) + C

グラフ

人気の例

d/(dy)((1+sin(y))/(1-sin(y)))

\frac{d}{d y} (\frac{1 + sin ( y )}{1 - sin ( y )})

integral of 20e^{-0.6t}

\int 20 e^{- 0.6 t} d t

integral from 0 to 4 of (2x+3)

\int_{0}^{4} (2 x + 3) d x

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

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

tangent f(x)=(x^3-9x)^{12},\at x=3

t an g e n t f (x) = (x^{3} - 9 x)^{12}, at x = 3