What is the integral of 1-6x+48x^2 ?

The integral of 1-6x+48x^2 is x-3x^2+16x^3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1-6x+48x^2

\int 1 - 6 x + 48 x^{2} d x

x - 3 x^{2} + 16 x^{3} + C

Solution steps

\int 1 - 6 x + 48 x^{2} 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 1 d x - \int 6 x d x + \int 48 x^{2} d x

\int 1 d x = x

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

\int 48 x^{2} d x = 16 x^{3}

= x - 3 x^{2} + 16 x^{3}

Add a constant to the solution

= x - 3 x^{2} + 16 x^{3} + C

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

\int 3 t^{2} + 1 d t

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

\int cos (4 x^{2}) d x

\int (x + x ln (x))^{- 1} d x

What is the integral of 1-6x+48x^2 ?
The integral of 1-6x+48x^2 is x-3x^2+16x^3+C