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 from 2 to 3 of x/(x^2+4x+20)

解

\int_{2}^{3} \frac{x}{x ^{2} + 4 x + 20} d x

解

\frac{1}{2} (ln (\frac{41}{16}) - ln (2) - arctan (\frac{5}{4}) + \frac{π}{4})

+1

十進法表記

0.06858 \dots

解答ステップ

\int_{2}^{3} \frac{x}{x ^{2} + 4 x + 20} d x

平方完成する

x^{2} + 4 x + 20 : (x + 2)^{2} + 16

= \int_{2}^{3} \frac{x}{( x + 2 ) ^{2} + 16} d x

u置換積分法を適用する

= \int_{4}^{5} \frac{u - 2}{u ^{2} + 16} d u

置換積分法を適用する

= \int_{1}^{\frac{5}{4}} \frac{2 v - 1}{2 ( v ^{2} + 1 )} d v

定数を除く:

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

= \frac{1}{2} \cdot \int_{1}^{\frac{5}{4}} \frac{2 v - 1}{v ^{2} + 1} d v

拡張

\frac{2 v - 1}{v ^{2} + 1} : \frac{2 v}{v ^{2} + 1} - \frac{1}{v ^{2} + 1}

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

= \frac{1}{2} (\int_{1}^{\frac{5}{4}} \frac{2 v}{v ^{2} + 1} d v - \int_{1}^{\frac{5}{4}} \frac{1}{v ^{2} + 1} d v)

\int_{1}^{\frac{5}{4}} \frac{2 v}{v ^{2} + 1} d v = ln (\frac{41}{16}) - ln (2)

\int_{1}^{\frac{5}{4}} \frac{1}{v ^{2} + 1} d v = arctan (\frac{5}{4}) - \frac{π}{4}

= \frac{1}{2} (ln (\frac{41}{16}) - ln (2) - (arctan (\frac{5}{4}) - \frac{π}{4}))

簡素化

= \frac{1}{2} (ln (\frac{41}{16}) - ln (2) - arctan (\frac{5}{4}) + \frac{π}{4})

グラフ

人気の例

(\partial)/(\partial x)((x^2)/(x+y))

\frac{\partial}{\partial x} (\frac{x ^{2}}{x + y})

derivative x*ln(x)

d er i v a t i v e x \cdot ln (x)

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

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

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

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

y^{''}-5y^'+4y=0,y(0)=-1/2 ,y^'(0)=-17/4

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = - \frac{1}{2}, y^{^{'}} (0) = - \frac{17}{4}