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受欢迎的微积分 >

integral of (sqrt(x))/(a+2x)

解答

\int \frac{x}{a + 2 x} d x

解答

- \frac{1}{2} a arctan (2 \frac{x}{a}) + x + C

求解步骤

\int \frac{x}{a + 2 x} d x

使用换元积分法

= \int \frac{u - a}{2 2 u} d u

提出常数:

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

= \frac{1}{2 2} \cdot \int \frac{u - a}{u} d u

使用换元积分法

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

提出常数:

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

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

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

= \frac{1}{2 2} \cdot 2 \cdot \int - \frac{a}{v ^{2} + a} + 1 d v

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{a}{v ^{2} + a} d v = a arctan (\frac{1}{a} v)

\int 1 d v = v

= \frac{1}{2 2} \cdot 2 (- a arctan (\frac{1}{a} v) + v)

代回

= \frac{1}{2 2} \cdot 2 (- a arctan (\frac{1}{a} a + 2 x - a) + a + 2 x - a)

化简

\frac{1}{2 2} \cdot 2 (- a arctan (\frac{1}{a} a + 2 x - a) + a + 2 x - a) : - \frac{1}{2} a arctan (2 \frac{x}{a}) + x

= - \frac{1}{2} a arctan (2 \frac{x}{a}) + x

解答补常数

= - \frac{1}{2} a arctan (2 \frac{x}{a}) + x + C

作图

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