What is the integral from 2 to 4 of (x^2+8)/(x^2) ?

The integral from 2 to 4 of (x^2+8)/(x^2) is 4

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integral from 2 to 4 of (x^2+8)/(x^2)

\int_{2}^{4} \frac{x ^{2} + 8}{x ^{2}} d x

4

Solution steps

\int_{2}^{4} \frac{x ^{2} + 8}{x ^{2}} d x

Expand

\frac{x ^{2} + 8}{x ^{2}} : 1 + \frac{8}{x ^{2}}

= \int_{2}^{4} 1 + \frac{8}{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

= \int_{2}^{4} 1 d x + \int_{2}^{4} \frac{8}{x ^{2}} d x

\int_{2}^{4} 1 d x = 2

\int_{2}^{4} \frac{8}{x ^{2}} d x = 2

= 2 + 2

Simplify

= 4

What is the integral from 2 to 4 of (x^2+8)/(x^2) ?
The integral from 2 to 4 of (x^2+8)/(x^2) is 4