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

The integral from 1 to x of 1/(t^3+1) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to x of 1/(t^3+1)

\int_{1}^{x} \frac{1}{t ^{3} + 1} d t

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

Solution steps

\int_{1}^{x} \frac{1}{t ^{3} + 1} d t

Take the partial fraction of

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

= \int_{1}^{x} \frac{1}{3 ( t + 1 )} + \frac{- t + 2}{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_{1}^{x} \frac{1}{3 ( t + 1 )} d t + \int_{1}^{x} \frac{- t + 2}{3 ( t ^{2} - t + 1 )} d t

\int_{1}^{x} \frac{1}{3 ( t + 1 )} d t = \frac{1}{3} (ln ∣ x + 1 ∣ - ln (2))

\int_{1}^{x} \frac{- t + 2}{3 ( t ^{2} - t + 1 )} d t = \frac{1}{3} (- \frac{1}{2} ln (\frac{4 x ^{2} - 4 x + 4}{4}) + 3 (arctan (\frac{2 x - 1}{3}) - \frac{π}{6}))

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

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

\int (\frac{x + 6}{x ^{2} + 11 x + 18}) d x

\frac{d}{d x} (x^{6} + 3 x^{4} + 7 x^{2})

cos (π t)

f^{^{''}} (x) = x^{\frac{- 3}{2}}, f^{^{'}} (4) = 2, f (0) = 0

What is the integral from 1 to x of 1/(t^3+1) ?
The integral from 1 to x of 1/(t^3+1) is