What is the integral of ((2x^3-x^2)^2)/(3x^3) ?

The integral of ((2x^3-x^2)^2)/(3x^3) is 1/3 (x^4-(4x^3)/3+(x^2)/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((2x^3-x^2)^2)/(3x^3)

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

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

Solution steps

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

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{( 2 x ^{3} - x ^{2} ) ^{2}}{x ^{3}} d x

Simplify

\frac{( 2 x ^{3} - x ^{2} ) ^{2}}{x ^{3}} : x (2 x - 1)^{2}

= \frac{1}{3} \cdot \int x (2 x - 1)^{2} d x

Expand

x (2 x - 1)^{2} : 4 x^{3} - 4 x^{2} + x

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

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

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

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

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

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

Add a constant to the solution

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

\int \frac{3}{x ^{2} + 4 x - 1} d x

\int \frac{10}{2 x + 5} d x

\int 5 e^{x} + 5 d x

\int e^{(- 4 x)} d x

What is the integral of ((2x^3-x^2)^2)/(3x^3) ?
The integral of ((2x^3-x^2)^2)/(3x^3) is 1/3 (x^4-(4x^3)/3+(x^2)/2)+C