What is the integral of x^2(x-8/x) ?

The integral of x^2(x-8/x) is (x^4)/4-4x^2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2(x-8/x)

\int x^{2} (x - \frac{8}{x}) d x

\frac{x ^{4}}{4} - 4 x^{2} + C

Solution steps

\int x^{2} (x - \frac{8}{x}) d x

Expand

x^{2} (x - \frac{8}{x}) : x^{3} - 8 x

= \int x^{3} - 8 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 x^{3} d x - \int 8 x d x

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

\int 8 x d x = 4 x^{2}

= \frac{x ^{4}}{4} - 4 x^{2}

Add a constant to the solution

= \frac{x ^{4}}{4} - 4 x^{2} + C

\int 2^{5 x} + 5 x^{2} - sin (5 x) + cos (5 π) d x

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

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

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

\int \frac{1}{3 x + x ln ( x )} d x

What is the integral of x^2(x-8/x) ?
The integral of x^2(x-8/x) is (x^4)/4-4x^2+C