What is the integral from 1 to 2 of (3/(x^2)-1) ?

The integral from 1 to 2 of (3/(x^2)-1) is 1/2

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integral from 1 to 2 of (3/(x^2)-1)

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

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

\frac{d y}{d x} = \frac{2 x + y + 1}{4 x + 2 y - 5}

x \to 0 lim (5 + 4 x - x^{3})

x \to - \infty lim (4)

x \to \infty lim (3 \cdot 2^{x})

\frac{\partial}{\partial y} (4 x y + 3 x^{2})

What is the integral from 1 to 2 of (3/(x^2)-1) ?
The integral from 1 to 2 of (3/(x^2)-1) is 1/2