What is the integral of t/(1+t^3) ?

The integral of t/(1+t^3) is -1/3 ln|t+1|+1/6 (ln|4t^2-4t+4|+2sqrt(3)arctan((2t-1)/(sqrt(3))))+C

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integral of t/(1+t^3)

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

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

Solution steps

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

Take the partial fraction of

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

Apply the Sum Rule:

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

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

Add a constant to the solution

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

What is the integral of t/(1+t^3) ?
The integral of t/(1+t^3) is -1/3 ln|t+1|+1/6 (ln|4t^2-4t+4|+2sqrt(3)arctan((2t-1)/(sqrt(3))))+C