What is the integral of (2x^6-3x^4-x^3)/(x^3) ?

The integral of (2x^6-3x^4-x^3)/(x^3) is (x^4)/2-(3x^2)/2-x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^6-3x^4-x^3)/(x^3)

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

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

Solution steps

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

Expand

\frac{2 x ^{6} - 3 x ^{4} - x ^{3}}{x ^{3}} : \frac{2 x ^{6}}{x ^{3}} - \frac{3 x ^{4}}{x ^{3}} - \frac{x ^{3}}{x ^{3}}

Apply the Sum Rule:

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

= \int \frac{2 x ^{6}}{x ^{3}} d x - \int \frac{3 x ^{4}}{x ^{3}} d x - \int \frac{x ^{3}}{x ^{3}} d x

\int \frac{2 x ^{6}}{x ^{3}} d x = \frac{x ^{4}}{2}

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

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

= \frac{x ^{4}}{2} - \frac{3 x ^{2}}{2} - x

Add a constant to the solution

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

\int (- 9 + \frac{8}{p}) d p

\int 2 z cos (z^{2}) d z

\int ln^{2} (x) x^{2} d x

\int 3 1 + x d x

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

What is the integral of (2x^6-3x^4-x^3)/(x^3) ?
The integral of (2x^6-3x^4-x^3)/(x^3) is (x^4)/2-(3x^2)/2-x+C