What is the integral of (5x\sqrt[3]{5x^2+1})/3 ?

The integral of (5x\sqrt[3]{5x^2+1})/3 is 1/8 (5x^2+1)^{4/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5x\sqrt[3]{5x^2+1})/3

\frac{1}{8} (5 x^{2} + 1)^{\frac{4}{3}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{3}{4} v^{\frac{4}{3}}

= \frac{5}{3} \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot \frac{3}{4} v^{\frac{4}{3}}

Substitute back

= \frac{5}{3} \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot \frac{3}{4} (5 x^{2} + 1)^{\frac{4}{3}}

Simplify

\frac{5}{3} \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot \frac{3}{4} (5 x^{2} + 1)^{\frac{4}{3}} : \frac{1}{8} (5 x^{2} + 1)^{\frac{4}{3}}

= \frac{1}{8} (5 x^{2} + 1)^{\frac{4}{3}}

Add a constant to the solution

= \frac{1}{8} (5 x^{2} + 1)^{\frac{4}{3}} + C

- cos^{2} (x)

\frac{8}{x - 2}

\int 6 sin (3 x) - cos (2 x) d x

\frac{d}{d x} (ln (sin (x^{2}) + 1))

\int 4 x^{6} - 2 x^{3} + 7 x - 4 d x

What is the integral of (5x\sqrt[3]{5x^2+1})/3 ?
The integral of (5x\sqrt[3]{5x^2+1})/3 is 1/8 (5x^2+1)^{4/3}+C