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

The integral of 1/(sqrt(x)(\sqrt[3]{x)+1)} is 6(-arctan(\sqrt[6]{x})+\sqrt[6]{x})+C

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

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

6 (- arctan (6 x) + 6 x) + C

Solution steps

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

Apply u-substitution

= \int \frac{6 u ^{2}}{u ^{2} + 1} d u

Take the constant out:

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

= 6 \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

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

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

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int 1 d u = u

= 6 (- arctan (u) + u)

Substitute back

u = 6 x

= 6 (- arctan (6 x) + 6 x)

Add a constant to the solution

= 6 (- arctan (6 x) + 6 x) + C

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What is the integral of 1/(sqrt(x)(\sqrt[3]{x)+1)} ?
The integral of 1/(sqrt(x)(\sqrt[3]{x)+1)} is 6(-arctan(\sqrt[6]{x})+\sqrt[6]{x})+C