What is the integral of 1/(x^3+2) ?

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integral of 1/(x^3+2)

\int \frac{1}{x ^{3} + 2} d x

Solution steps

\int \frac{1}{x ^{3} + 2} d x

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

= \int \frac{1}{2 ( \frac{x ^{3}}{2} + 1 )} d x

Apply exponent rule:

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

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Take the partial fraction of

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

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 ( u + 1 )} d u = \frac{1}{3} ln ∣ u + 1 ∣

\int \frac{- u + 2}{3 ( u ^{2} - u + 1 )} d u = \frac{1}{6} (- ln \frac{4}{3} + \frac{4 u ^{2} - 4 u}{3} + 23 arctan (\frac{2 u - 1}{3}))

Substitute back

Simplify

Add a constant to the solution

f (x) = x^{2} - 3, (3, 6)

\frac{d y}{d x} = 3 x e^{y}, y (0) = 0

\int_{1}^{7} \frac{2 ln ( x )}{x} d x

\int_{0}^{1} 6^{4 x} d x

y^{^{''}} - 3 y^{^{'}} = 18 t