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受欢迎的微积分 >

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

解答

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

解答

\frac{4}{5} x^{\frac{5}{4}} - x + \frac{4}{3} x^{\frac{3}{4}} - 2 x + 4 4 x - 4 ln 4 x + 1 + C

求解步骤

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

使用换元积分法

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

使用换元积分法

= 2 \cdot \int \frac{2 v ^{5}}{v + 1} 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 ^{5}}{v + 1} d v

长除

\frac{v ^{5}}{v + 1} : v^{4} - v^{3} + v^{2} - v + 1 - \frac{1}{v + 1}

= 2 \cdot 2 \cdot \int v^{4} - v^{3} + v^{2} - v + 1 - \frac{1}{v + 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

= 2 \cdot 2 (\int v^{4} d v - \int v^{3} d v + \int v^{2} d v - \int v d v + \int 1 d v - \int \frac{1}{v + 1} d v)

\int v^{4} d v = \frac{v ^{5}}{5}

\int v^{3} d v = \frac{v ^{4}}{4}

\int v^{2} d v = \frac{v ^{3}}{3}

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

\int 1 d v = v

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

= 2 \cdot 2 (\frac{v ^{5}}{5} - \frac{v ^{4}}{4} + \frac{v ^{3}}{3} - \frac{v ^{2}}{2} + v - ln ∣ v + 1 ∣)

代回

= 2 \cdot 2 \frac{( x ) ^{5}}{5} - \frac{( x ) ^{4}}{4} + \frac{( x ) ^{3}}{3} - \frac{( x ) ^{2}}{2} + x - ln x + 1

化简

2 \cdot 2 \frac{( x ) ^{5}}{5} - \frac{( x ) ^{4}}{4} + \frac{( x ) ^{3}}{3} - \frac{( x ) ^{2}}{2} + x - ln x + 1 : \frac{4}{5} x^{\frac{5}{4}} - x + \frac{4}{3} x^{\frac{3}{4}} - 2 x + 4 4 x - 4 ln 4 x + 1

= \frac{4}{5} x^{\frac{5}{4}} - x + \frac{4}{3} x^{\frac{3}{4}} - 2 x + 4 4 x - 4 ln 4 x + 1

解答补常数

= \frac{4}{5} x^{\frac{5}{4}} - x + \frac{4}{3} x^{\frac{3}{4}} - 2 x + 4 4 x - 4 ln 4 x + 1 + C

作图

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