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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2-4t-6t^3)

\int (2 - 4 t - 6 t^{3}) d t

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

Solution steps

\int 2 - 4 t - 6 t^{3} d t

Apply the Sum Rule:

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

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

\int 2 d t = 2 t

\int 4 t d t = 2 t^{2}

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

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

Add a constant to the solution

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

\int 5 x \cdot cos (x) d x

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

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

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

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