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

解答

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

6 (- arctan (6 x) + 6 x) + C

求解步骤

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

使用换元积分法

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

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

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

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

\int 1 d u = u

= 6 (- arctan (u) + u)

u = 6 x

代回

= 6 (- arctan (6 x) + 6 x)

解答补常数

= 6 (- arctan (6 x) + 6 x) + C

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