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+16))/x

解

\int \frac{x + 16}{x} d x

解

- 8 (\frac{1}{2} ln \frac{x + 16}{4} + 1 - \frac{1}{2} ln \frac{x + 16}{4} - 1 - \frac{1}{4} x + 16) + C

解答ステップ

\int \frac{x + 16}{x} d x

u置換積分法を適用する

= \int \frac{2 u ^{2}}{u ^{2} - 16} 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{u ^{2}}{u ^{2} - 16} d u

因数

u^{2} - 16 : - (- u^{2} + 16)

= 2 \cdot \int \frac{u ^{2}}{- ( - u ^{2} + 16 )} d u

定数を除く:

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

= 2 (- \int \frac{u ^{2}}{- u ^{2} + 16} d u)

置換積分法を適用する

= 2 (- \int \frac{4 v ^{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

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

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

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

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 (- 4 (\int \frac{1}{- v ^{2} + 1} d v - \int 1 d v))

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

\int 1 d v = v

= 2 (- 4 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v))

代用を戻す

= 2 - 4 \frac{ln \frac{x + 16}{4} + 1}{2} - \frac{ln \frac{x + 16}{4} - 1}{2} - \frac{x + 16}{4}

簡素化

2 - 4 \frac{ln \frac{x + 16}{4} + 1}{2} - \frac{ln \frac{x + 16}{4} - 1}{2} - \frac{x + 16}{4} : - 8 (\frac{1}{2} ln \frac{x + 16}{4} + 1 - \frac{1}{2} ln \frac{x + 16}{4} - 1 - \frac{1}{4} x + 16)

= - 8 (\frac{1}{2} ln \frac{x + 16}{4} + 1 - \frac{1}{2} ln \frac{x + 16}{4} - 1 - \frac{1}{4} x + 16)

定数を解答に追加する

= - 8 (\frac{1}{2} ln \frac{x + 16}{4} + 1 - \frac{1}{2} ln \frac{x + 16}{4} - 1 - \frac{1}{4} x + 16) + C

グラフ

人気の例

limit as x approaches-3 of-(3x)

x \to - 3 lim (- (3 x))

derivative e^{(3h^2+6)}

d er i v a t i v e e^{(3 h^{2} + 6)}

limit as t approaches 0 of ((e^t-1))/t

t \to 0 lim (\frac{( e ^{t} - 1 )}{t})

derivative of (x+y^2+(x+1)^2)

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

inverselaplace (-20)/(s(s+10))

in v erse l a pl a ce \frac{- 20}{s ( s + 10 )}