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

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

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

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

1

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{1}^{2} 1 - \frac{1}{x ^{2}} 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

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

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

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

= 2 (1 - \frac{1}{2})

Simplify

2 (1 - \frac{1}{2}) : 1

= 1

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

f (x) = \frac{x}{x ^{2} + 7}

\frac{d}{d x} (\frac{3}{4 ( 1 - x ) ^{\frac{5}{2}}})

4 x^{2} y^{^{''}} + y = 0

\int \frac{x ^{2}}{6} d x

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