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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (t+1)(t-1)

\int (t + 1) (t - 1) d t

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

Solution steps

\int (t + 1) (t - 1) d t

Expand

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

= \int t^{2} - 1 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 t^{2} d t - \int 1 d t

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

\int 1 d t = t

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

Add a constant to the solution

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

\int \frac{6 x ^{4} + x ^{3} - 2 x - 1}{x ^{2}} d x

\int \frac{2 ^{2 x} - 1}{2 ^{x}} d x

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

\int \frac{sin ( ln ( 15 x ) )}{x} d x

\int \frac{21 x + 1}{( 21 x ^{2} + 2 x ) ^{3}} d x