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受欢迎的微积分 >

integral of 1/((sqrt(x)+\sqrt[4]{x))}

解答

\int \frac{1}{( x + 4 x )} d x

解答

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

求解步骤

\int \frac{1}{x + 4 x} d x

使用换元积分法

= \int \frac{2 u}{u + 1} d u

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

乘开

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

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

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

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

\int 2 d v = 2 v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 \cdot 2 (\frac{v ^{2}}{2} - 2 v + ln ∣ v ∣)

代回

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

化简

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

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

解答补常数

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

作图

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