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受欢迎的微积分 >

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

解答

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

解答

3 3 y + 1 + ln 3 y + 1 - 1 - \frac{1}{2} ln 4 (y + 1)^{\frac{2}{3}} + 4 3 y + 1 + 4 - 3 arctan (\frac{1}{3} (2 3 y + 1 + 1)) + C

求解步骤

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

使用换元积分法

= \int \frac{3 u ^{3}}{u ^{3} - 1} d u

提出常数:

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

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

将

\frac{u ^{3}}{u ^{3} - 1}

用部份分式展开:

1 + \frac{1}{3 ( u - 1 )} + \frac{- u - 2}{3 ( u ^{2} + u + 1 )}

= 3 \cdot \int 1 + \frac{1}{3 ( u - 1 )} + \frac{- u - 2}{3 ( u ^{2} + u + 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

= 3 (\int 1 d u + \int \frac{1}{3 ( u - 1 )} d u + \int \frac{- u - 2}{3 ( u ^{2} + u + 1 )} d u)

\int 1 d u = u

\int \frac{1}{3 ( u - 1 )} d u = \frac{1}{3} ln ∣ u - 1 ∣

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

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

u = 3 y + 1

代回

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

化简

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

= 3 3 y + 1 + ln 3 y + 1 - 1 - \frac{1}{2} ln 4 (y + 1)^{\frac{2}{3}} + 4 3 y + 1 + 4 - 3 arctan (\frac{1}{3} (2 3 y + 1 + 1))

解答补常数

= 3 3 y + 1 + ln 3 y + 1 - 1 - \frac{1}{2} ln 4 (y + 1)^{\frac{2}{3}} + 4 3 y + 1 + 4 - 3 arctan (\frac{1}{3} (2 3 y + 1 + 1)) + C

作图

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