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受欢迎的微积分 >

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

解答

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

解答

6 (\frac{1}{7} x^{\frac{7}{6}} - \frac{1}{5} x^{\frac{5}{6}} + \frac{1}{3} x - 6 x + arctan (6 x)) + C

求解步骤

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

使用换元积分法

= \int \frac{6 u ^{8}}{1 + u ^{2}} 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 ^{8}}{1 + u ^{2}} d u

长除

\frac{u ^{8}}{1 + u ^{2}} : u^{6} - u^{4} + u^{2} - 1 + \frac{1}{u ^{2} + 1}

= 6 \cdot \int u^{6} - u^{4} + u^{2} - 1 + \frac{1}{u ^{2} + 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^{6} d u - \int u^{4} d u + \int u^{2} d u - \int 1 d u + \int \frac{1}{u ^{2} + 1} d u)

\int u^{6} d u = \frac{u ^{7}}{7}

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

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

\int 1 d u = u

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

= 6 (\frac{u ^{7}}{7} - \frac{u ^{5}}{5} + \frac{u ^{3}}{3} - u + arctan (u))

u = 6 x

代回

= 6 (\frac{( 6 x ) ^{7}}{7} - \frac{( 6 x ) ^{5}}{5} + \frac{( 6 x ) ^{3}}{3} - 6 x + arctan (6 x))

化简

6 (\frac{( 6 x ) ^{7}}{7} - \frac{( 6 x ) ^{5}}{5} + \frac{( 6 x ) ^{3}}{3} - 6 x + arctan (6 x)) : 6 (\frac{1}{7} x^{\frac{7}{6}} - \frac{1}{5} x^{\frac{5}{6}} + \frac{1}{3} x - 6 x + arctan (6 x))

= 6 (\frac{1}{7} x^{\frac{7}{6}} - \frac{1}{5} x^{\frac{5}{6}} + \frac{1}{3} x - 6 x + arctan (6 x))

解答补常数

= 6 (\frac{1}{7} x^{\frac{7}{6}} - \frac{1}{5} x^{\frac{5}{6}} + \frac{1}{3} x - 6 x + arctan (6 x)) + C

作图

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