What is the integral from 1 to 2 of (x^2-12)/(x^4) ?

The integral from 1 to 2 of (x^2-12)/(x^4) is -3

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integral from 1 to 2 of (x^2-12)/(x^4)

\int_{1}^{2} \frac{x ^{2} - 12}{x ^{4}} d x

- 3

Solution steps

\int_{1}^{2} \frac{x ^{2} - 12}{x ^{4}} d x

Expand

\frac{x ^{2} - 12}{x ^{4}} : \frac{1}{x ^{2}} - \frac{12}{x ^{4}}

= \int_{1}^{2} \frac{1}{x ^{2}} - \frac{12}{x ^{4}} 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{1}{x ^{2}} d x - \int_{1}^{2} \frac{12}{x ^{4}} d x

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

\int_{1}^{2} \frac{12}{x ^{4}} d x = \frac{7}{2}

= \frac{1}{2} - \frac{7}{2}

Simplify

\frac{1}{2} - \frac{7}{2} : - 3

= - 3

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What is the integral from 1 to 2 of (x^2-12)/(x^4) ?
The integral from 1 to 2 of (x^2-12)/(x^4) is -3