What is the integral from 1 to 2 of (x^2-1)/(x+1) ?

The integral from 1 to 2 of (x^2-1)/(x+1) is 1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of (x^2-1)/(x+1)

\int_{1}^{2} \frac{x ^{2} - 1}{x + 1} d x

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{1}^{2} \frac{x ^{2} - 1}{x + 1} d x

Simplify

\frac{x ^{2} - 1}{x + 1} : x - 1

= \int_{1}^{2} x - 1 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}^{2} x d x - \int_{1}^{2} 1 d x

\int_{1}^{2} x d x = \frac{3}{2}

\int_{1}^{2} 1 d x = 1

= \frac{3}{2} - 1

Simplify

\frac{3}{2} - 1 : \frac{1}{2}

= \frac{1}{2}

y = x^{9 x}

\int \frac{( ln ( x ) ) ^{3}}{3 x} d x

n = 0 \sum \infty \frac{n x ^{n}}{n + 8}

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

\frac{\partial}{\partial x} (2 x^{6 y})

What is the integral from 1 to 2 of (x^2-1)/(x+1) ?
The integral from 1 to 2 of (x^2-1)/(x+1) is 1/2