What is the integral from 0 to x of 1/(2+t^3) ?

The integral from 0 to x of 1/(2+t^3) is

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integral from 0 to x of 1/(2+t^3)

\int_{0}^{x} \frac{1}{2 + t ^{3}} d t

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

Solution steps

\int_{0}^{x} \frac{1}{2 + t ^{3}} d t

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

= \int_{0}^{x} \frac{1}{2 ( 1 + \frac{t ^{3}}{2} )} d t

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{t ^{3}}{2} = (\frac{t}{3 2})^{3}

= \int_{0}^{x} \frac{1}{2 ( 1 + ( \frac{t}{3 2} ) ^{3} )} d t

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{x} \frac{1}{1 + ( \frac{t}{3 2} ) ^{3}} d t

Apply u-substitution

= \frac{1}{2} \cdot \int_{0}^{\frac{x}{3 2}} \frac{3 2}{1 + u ^{3}} d u

Take the constant out:

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

= \frac{1}{2} 32 \cdot \int_{0}^{\frac{x}{3 2}} \frac{1}{1 + u ^{3}} d u

Take the partial fraction of

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

= \frac{1}{2} 32 \cdot \int_{0}^{\frac{x}{3 2}} \frac{1}{3 ( u + 1 )} + \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} 32 (\int_{0}^{\frac{x}{3 2}} \frac{1}{3 ( u + 1 )} d u + \int_{0}^{\frac{x}{3 2}} \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u)

\int_{0}^{\frac{x}{3 2}} \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln \frac{x}{3 2} + 1

\int_{0}^{\frac{x}{3 2}} \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u = \frac{- ln \frac{3 \frac{2 3 2 x ^{2} - 2 ^{1 + \frac{2}{3}} x + 1}{3} + 1}{4} + 2 3 ( arctan ( \frac{2 ^{\frac{2}{3}} x - 1}{3} ) + \frac{π}{6} )}{6}

= \frac{1}{2} 32 \frac{1}{3} ln \frac{x}{3 2} + 1 + \frac{- ln \frac{3 \frac{2 3 2 x ^{2} - 2 ^{1 + \frac{2}{3}} x + 1}{3} + 1}{4} + 2 3 ( arctan ( \frac{2 ^{\frac{2}{3}} x - 1}{3} ) + \frac{π}{6} )}{6}

Simplify

\frac{1}{2} 32 \frac{1}{3} ln \frac{x}{3 2} + 1 + \frac{- ln \frac{3 \frac{2 3 2 x ^{2} - 2 ^{1 + \frac{2}{3}} x + 1}{3} + 1}{4} + 2 3 ( arctan ( \frac{2 ^{\frac{2}{3}} x - 1}{3} ) + \frac{π}{6} )}{6} : \frac{2 ln \frac{x}{3 2} + 1 + 2 3 ( arctan ( \frac{2 ^{\frac{2}{3}} x - 1}{3} ) + \frac{π}{6} ) - ln \frac{3 \frac{2 3 2 x ^{2} - 2 ^{1 + \frac{2}{3}} x + 1}{3} + 1}{4}}{6 \cdot 2 ^{\frac{2}{3}}}

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

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What is the integral from 0 to x of 1/(2+t^3) ?
The integral from 0 to x of 1/(2+t^3) is