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受欢迎的微积分 >

integral of (\sqrt[3]{x})/(1+sqrt(x))

解答

\int \frac{3 x}{1 + x} d x

解答

\frac{6}{5} x^{\frac{5}{6}} - 3 3 x - 2 ln 6 x + 1 + ln 4 3 x - 4 6 x + 4 + 23 arctan (\frac{1}{3} (2 6 x - 1)) + C

求解步骤

\int \frac{3 x}{1 + x} d x

使用换元积分法

= \int \frac{6 u ^{7}}{1 + u ^{3}} d u

提出常数:

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

= 6 \cdot \int \frac{u ^{7}}{1 + u ^{3}} d u

长除

\frac{u ^{7}}{1 + u ^{3}} : u^{4} - u + \frac{u}{u ^{3} + 1}

= 6 \cdot \int u^{4} - u + \frac{u}{u ^{3} + 1} d u

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

= 6 (\int u^{4} d u - \int u d u + \int \frac{u}{u ^{3} + 1} d u)

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

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

\int \frac{u}{u ^{3} + 1} d u = - \frac{1}{3} ln ∣ u + 1 ∣ + \frac{1}{6} (ln 4 u^{2} - 4 u + 4 + 23 arctan (\frac{2 u - 1}{3}))

= 6 (\frac{u ^{5}}{5} - \frac{u ^{2}}{2} - \frac{1}{3} ln ∣ u + 1 ∣ + \frac{1}{6} (ln 4 u^{2} - 4 u + 4 + 23 arctan (\frac{2 u - 1}{3})))

u = 6 x

代回

= 6 (\frac{( 6 x ) ^{5}}{5} - \frac{( 6 x ) ^{2}}{2} - \frac{1}{3} ln 6 x + 1 + \frac{1}{6} (ln 4 (6 x)^{2} - 4 6 x + 4 + 23 arctan (\frac{2 6 x - 1}{3})))

化简

6 (\frac{( 6 x ) ^{5}}{5} - \frac{( 6 x ) ^{2}}{2} - \frac{1}{3} ln 6 x + 1 + \frac{1}{6} (ln 4 (6 x)^{2} - 4 6 x + 4 + 23 arctan (\frac{2 6 x - 1}{3}))) : \frac{6}{5} x^{\frac{5}{6}} - 3 3 x - 2 ln 6 x + 1 + ln 4 3 x - 4 6 x + 4 + 23 arctan (\frac{1}{3} (2 6 x - 1))

= \frac{6}{5} x^{\frac{5}{6}} - 3 3 x - 2 ln 6 x + 1 + ln 4 3 x - 4 6 x + 4 + 23 arctan (\frac{1}{3} (2 6 x - 1))

解答补常数

= \frac{6}{5} x^{\frac{5}{6}} - 3 3 x - 2 ln 6 x + 1 + ln 4 3 x - 4 6 x + 4 + 23 arctan (\frac{1}{3} (2 6 x - 1)) + C

作图

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