What is the integral of ((x^2)/3-8x-3) ?

The integral of ((x^2)/3-8x-3) is (x^3)/9-4x^2-3x+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)/3-8x-3)

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

\frac{x ^{3}}{9} - 4 x^{2} - 3 x + C

Solution steps

\int \frac{x ^{2}}{3} - 8 x - 3 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 \frac{x ^{2}}{3} d x - \int 8 x d x - \int 3 d x

\int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{9}

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

\int 3 d x = 3 x

= \frac{x ^{3}}{9} - 4 x^{2} - 3 x

Add a constant to the solution

= \frac{x ^{3}}{9} - 4 x^{2} - 3 x + C

\int sin^{5} (x) cos^{9} (x) d x

\int 16 x^{4} - 24 x^{2} + 9 d x

\int \frac{1}{( 7 - x ^{2} ) ^{\frac{3}{2}}} d x

\int \frac{x + 1}{x ^{2} + 10 x + 18} d x

\int \frac{1}{3 x ^{2} - 2 x + 5} d x

What is the integral of ((x^2)/3-8x-3) ?
The integral of ((x^2)/3-8x-3) is (x^3)/9-4x^2-3x+C