What is the integral from 1 to 4 of (3/(x^2)-1) ?

The integral from 1 to 4 of (3/(x^2)-1) is -3/4

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integral from 1 to 4 of (3/(x^2)-1)

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

- \frac{3}{4}

Decimal Notation

- 0.75

Solution steps

\int_{1}^{4} \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}^{4} \frac{3}{x ^{2}} d x - \int_{1}^{4} 1 d x

\int_{1}^{4} \frac{3}{x ^{2}} d x = \frac{9}{4}

\int_{1}^{4} 1 d x = 3

= \frac{9}{4} - 3

Simplify

\frac{9}{4} - 3 : - \frac{3}{4}

= - \frac{3}{4}

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What is the integral from 1 to 4 of (3/(x^2)-1) ?
The integral from 1 to 4 of (3/(x^2)-1) is -3/4