What is the integral of (1-t)(6+t^2) ?

The integral of (1-t)(6+t^2) is 6t+(t^3)/3-3t^2-(t^4)/4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1-t)(6+t^2)

\int (1 - t) (6 + t^{2}) d t

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

Solution steps

\int (1 - t) (6 + t^{2}) d t

Expand

(1 - t) (6 + t^{2}) : 6 + t^{2} - 6 t - t^{3}

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

\int 6 d t = 6 t

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

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

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

= 6 t + \frac{t ^{3}}{3} - 3 t^{2} - \frac{t ^{4}}{4}

Add a constant to the solution

= 6 t + \frac{t ^{3}}{3} - 3 t^{2} - \frac{t ^{4}}{4} + C

\int \frac{10 x}{4 + 25 x ^{4}} d x

\int \frac{1}{9 x ^{5}} d x

\int x^{8} + sec^{2} (x) d x

\int 2 cos (2 x) sin (2 x) d x

\int 2 tan (t) d t

What is the integral of (1-t)(6+t^2) ?
The integral of (1-t)(6+t^2) is 6t+(t^3)/3-3t^2-(t^4)/4+C