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人気のある微分積分 >

integral of 1/(sqrt(x+1)+x-1)

解

\int \frac{1}{x + 1 + x - 1} d x

解

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

解答ステップ

\int \frac{1}{x + 1 + x - 1} d x

u置換積分法を適用する

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

以下の部分分数を得る:

\frac{u}{u ^{2} + u - 2} : \frac{1}{3 ( u - 1 )} + \frac{2}{3 ( u + 2 )}

= 2 \cdot \int \frac{1}{3 ( u - 1 )} + \frac{2}{3 ( u + 2 )} 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

= 2 (\int \frac{1}{3 ( u - 1 )} d u + \int \frac{2}{3 ( u + 2 )} d u)

\int \frac{1}{3 ( u - 1 )} d u = \frac{1}{3} ln ∣ u - 1 ∣

\int \frac{2}{3 ( u + 2 )} d u = \frac{2}{3} ln ∣ u + 2 ∣

= 2 (\frac{1}{3} ln ∣ u - 1 ∣ + \frac{2}{3} ln ∣ u + 2 ∣)

代用を戻す

u = x + 1

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

定数を解答に追加する

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

グラフ

人気の例

integral of 4/((4+x^2)^2)

\int \frac{4}{( 4 + x ^{2} ) ^{2}} d x

limit as x approaches 1+of x^{4/(1-x)}

x \to 1 + lim (x^{\frac{4}{1 - x}})

limit as x approaches 1+of 4/(x-1)

x \to 1 + lim (\frac{4}{x - 1})

(\partial)/(\partial x)(2x^2y-2x+4y^3)

\frac{\partial}{\partial x} (2 x^{2} y - 2 x + 4 y^{3})

limit as x approaches 0 of x^{2sqrt(x)}

x \to 0 lim (x^{2 x})