What is the integral of 1/(x(4+\sqrt[3]{x))} ?

The integral of 1/(x(4+\sqrt[3]{x))} is 3(1/4 ln|\sqrt[3]{x}|-1/4 ln|\sqrt[3]{x}+4|)+C

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integral of 1/(x(4+\sqrt[3]{x))}

Solution steps

Apply u-substitution

: \int \frac{3}{u ( u + 4 )} d u

= \int \frac{3}{u ( u + 4 )} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{u ( u + 4 )} d u

Take the partial fraction of

\frac{1}{u ( u + 4 )} : \frac{1}{4 u} - \frac{1}{4 ( u + 4 )}

= 3 \cdot \int \frac{1}{4 u} - \frac{1}{4 ( u + 4 )} 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 \frac{1}{4 u} d u - \int \frac{1}{4 ( u + 4 )} d u)

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{4 ( u + 4 )} d u = \frac{1}{4} ln ∣ u + 4 ∣

= 3 (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 4 ∣)

Substitute back

Add a constant to the solution

\int \frac{x ^{2} + 3 x + 2}{x ^{3} + 4 x} d x

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\int \frac{4 x ^{3} - 8 x ^{2} + 4 x + 3}{4 x ^{2} - 8 x + 5} d x

\int (5 x^{4} - 3 x^{4} + 4 x^{2} - 2 + 6) d x

What is the integral of 1/(x(4+\sqrt[3]{x))} ?
The integral of 1/(x(4+\sqrt[3]{x))} is 3(1/4 ln|\sqrt[3]{x}|-1/4 ln|\sqrt[3]{x}+4|)+C