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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{x}{3 1 + 2 x} d x

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

Solution steps

\int \frac{x}{3 1 + 2 x} d x

Apply u-substitution

: \int \frac{u - 1}{4 3 u} d u

= \int \frac{u - 1}{4 3 u} d u

Take the constant out:

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

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

Expand

\frac{u - 1}{3 u} : u^{\frac{2}{3}} - \frac{1}{3 u}

= \frac{1}{4} \cdot \int u^{\frac{2}{3}} - \frac{1}{3 u} 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

= \frac{1}{4} (\int u^{\frac{2}{3}} d u - \int \frac{1}{3 u} d u)

\int u^{\frac{2}{3}} d u = \frac{3}{5} u^{\frac{5}{3}}

\int \frac{1}{3 u} d u = \frac{3}{2} u^{\frac{2}{3}}

= \frac{1}{4} (\frac{3}{5} u^{\frac{5}{3}} - \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = 1 + 2 x

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

Add a constant to the solution

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

\int sin^{3 x} (θ) d y

\int d arcsin (x) d x

\int \frac{x ^{3} + 4 x ^{2} - 25 x - 12}{x - 3} d x

\int \frac{5 - θ ^{2}}{θ ^{2}} d θ

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

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