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受欢迎的微积分 >

integral of t/(1-t^3)

解答

\int \frac{t}{1 - t ^{3}} d t

解答

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

求解步骤

\int \frac{t}{1 - t ^{3}} d t

因式分解

1 - t^{3} : - (- 1 + t^{3})

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

提出常数:

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

= - \int \frac{t}{- 1 + t ^{3}} d t

将

\frac{t}{- 1 + t ^{3}}

用部份分式展开:

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

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

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

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

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

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

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

化简

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

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

解答补常数

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

作图

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