What is the integral of \sqrt[3]{x}+10\sqrt[5]{x} ?

The integral of \sqrt[3]{x}+10\sqrt[5]{x} is 3/4 x^{4/3}+25/3 x^{6/5}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of \sqrt[3]{x}+10\sqrt[5]{x}

\int 3 x + 10 5 x d x

\frac{3}{4} x^{\frac{4}{3}} + \frac{25}{3} x^{\frac{6}{5}} + C

Solution steps

\int 3 x + 10 5 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 x d x + \int 10 5 x d x

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

\int 10 5 x d x = \frac{25}{3} x^{\frac{6}{5}}

= \frac{3}{4} x^{\frac{4}{3}} + \frac{25}{3} x^{\frac{6}{5}}

Add a constant to the solution

= \frac{3}{4} x^{\frac{4}{3}} + \frac{25}{3} x^{\frac{6}{5}} + C

\frac{\partial}{\partial x} (e^{\frac{y}{x}})

\int x^{2} + 3 d x

\int_{0}^{1} \frac{2}{3} x^{2} (x + 1) d x

\int 4 (tan^{2} (x) + tan^{4} (x)) d x

y^{^{''}} - 6 y^{^{'}} + 5 y = 0

What is the integral of \sqrt[3]{x}+10\sqrt[5]{x} ?
The integral of \sqrt[3]{x}+10\sqrt[5]{x} is 3/4 x^{4/3}+25/3 x^{6/5}+C