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

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

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

Solution steps

Apply u-substitution

: \int \frac{12 u ^{8}}{u + 1} d u

= \int \frac{12 u ^{8}}{u + 1} d u

Take the constant out:

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

= 12 \cdot \int \frac{u ^{8}}{u + 1} d u

Long division

\frac{u ^{8}}{u + 1} : u^{7} - u^{6} + u^{5} - u^{4} + u^{3} - u^{2} + u - 1 + \frac{1}{u + 1}

= 12 \cdot \int u^{7} - u^{6} + u^{5} - u^{4} + u^{3} - u^{2} + u - 1 + \frac{1}{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

= 12 (\int u^{7} d u - \int u^{6} d u + \int u^{5} d u - \int u^{4} d u + \int u^{3} d u - \int u^{2} d u + \int u d u - \int 1 d u + \int \frac{1}{u + 1} d u)

\int u^{7} d u = \frac{u ^{8}}{8}

\int u^{6} d u = \frac{u ^{7}}{7}

\int u^{5} d u = \frac{u ^{6}}{6}

\int u^{4} d u = \frac{u ^{5}}{5}

\int u^{3} d u = \frac{u ^{4}}{4}

\int u^{2} d u = \frac{u ^{3}}{3}

\int u d u = \frac{u ^{2}}{2}

\int 1 d u = u

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

= 12 (\frac{u ^{8}}{8} - \frac{u ^{7}}{7} + \frac{u ^{6}}{6} - \frac{u ^{5}}{5} + \frac{u ^{4}}{4} - \frac{u ^{3}}{3} + \frac{u ^{2}}{2} - u + ln ∣ u + 1 ∣)

Substitute back

Simplify

Add a constant to the solution

x \to 0 - lim (\frac{6}{5 x ^{\frac{1}{3}}})

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

x \to \infty lim (5 x + \frac{1}{x})

\frac{\partial}{\partial x} (t^{2} + 4 x^{3})

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