What is the integral of (\sqrt[3]{x+27})/x ?

The integral of (\sqrt[3]{x+27})/x is

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integral of (\sqrt[3]{x+27})/x

Solution steps

Apply u-substitution

: \int \frac{3 u ^{3}}{u ^{3} - 27} d u

= \int \frac{3 u ^{3}}{u ^{3} - 27} d u

Take the constant out:

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

= 3 \cdot \int \frac{u ^{3}}{u ^{3} - 27} d u

Take the partial fraction of

\frac{u ^{3}}{u ^{3} - 27} : 1 + \frac{1}{u - 3} + \frac{- u - 6}{u ^{2} + 3 u + 9}

= 3 \cdot \int 1 + \frac{1}{u - 3} + \frac{- u - 6}{u ^{2} + 3 u + 9} 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

= 3 (\int 1 d u + \int \frac{1}{u - 3} d u + \int \frac{- u - 6}{u ^{2} + 3 u + 9} d u)

\int 1 d u = u

\int \frac{1}{u - 3} d u = ln ∣ u - 3 ∣

\int \frac{- u - 6}{u ^{2} + 3 u + 9} d u = - \frac{1}{2} ln 4 u^{2} + 12 u + 36 - 3 arctan (\frac{2}{3 3} u + \frac{1}{3})

= 3 (u + ln ∣ u - 3 ∣ - \frac{1}{2} ln 4 u^{2} + 12 u + 36 - 3 arctan (\frac{2}{3 3} u + \frac{1}{3}))

Substitute back

Add a constant to the solution

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

\int 9^{x} e^{2 x} d x

\int (x - 2) (x + 3)^{2} d x

\int (2 x - 8) x - 4 d x

\int (x^{2} + 2 x) e^{2 x} d x

What is the integral of (\sqrt[3]{x+27})/x ?
The integral of (\sqrt[3]{x+27})/x is