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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int 5 x e^{- 3 x^{2}} d x

\int \frac{5 x ^{2} + 2}{x ^{3} - 4 x ^{2} + 5 x} d x

\int \frac{d}{1} (x y) d x

\int x^{2} 3 x^{3} - 1 d x

\int \frac{( x ^{2} + 1 )}{x} d x

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