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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

定数を除く:

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

= \frac{1}{2} \cdot \int \frac{x ^{2}}{x + 1} d x

u置換積分法を適用する

= \frac{1}{2} \cdot \int \frac{( u - 1 ) ^{2}}{u} d u

拡張

\frac{( u - 1 ) ^{2}}{u} : u - 2 + \frac{1}{u}

= \frac{1}{2} \cdot \int u - 2 + \frac{1}{u} 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

= \frac{1}{2} (\int u d u - \int 2 d u + \int \frac{1}{u} d u)

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{2} (\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣)

代用を戻す

u = x + 1

= \frac{1}{2} (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣)

簡素化

\frac{1}{2} (\frac{( x + 1 ) ^{2}}{2} - 2 (x + 1) + ln ∣ x + 1 ∣) : \frac{1}{4} (x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3)

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

定数を解答に追加する

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

グラフ

人気の例

derivative of e^{9x^2}

\frac{d}{d x} (e^{9 x^{2}})

x^{^{''}} + 2 x = 0

derivative y=x^3-3x+1

d er i v a t i v e y = x^{3} - 3 x + 1

(\partial)/(\partial x)(sin(x)+x^2e^y-1)

\frac{\partial}{\partial x} (sin (x) + x^{2} e^{y} - 1)

derivative f(x)=(ln(x))/(x^3)

d er i v a t i v e f (x) = \frac{ln ( x )}{x ^{3}}