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

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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}))

\int \frac{1}{sec ( a x ) + 1} d x

\frac{d y}{d x} = \frac{( ( 4 x ) sec ( \frac{y}{x} ) + y )}{x}

y^{^{'}} = \frac{( y ^{2} )}{x ^{3}}

\frac{\partial}{\partial x} (3 x^{2} y^{2} + 1)

\int \frac{1}{10 x - 3} d x

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