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

The integral from 1 to 2 of (x^2+x^4)/(x^2) is 10/3

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

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

\frac{10}{3}

Decimal

3.33333 \dots

Solution steps

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

Expand

\frac{x ^{2} + x ^{4}}{x ^{2}} : \frac{x ^{2}}{x ^{2}} + \frac{x ^{4}}{x ^{2}}

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

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

\int_{1}^{2} \frac{x ^{4}}{x ^{2}} d x = \frac{7}{3}

= 1 + \frac{7}{3}

Simplify

1 + \frac{7}{3} : \frac{10}{3}

= \frac{10}{3}

What is the integral from 1 to 2 of (x^2+x^4)/(x^2) ?
The integral from 1 to 2 of (x^2+x^4)/(x^2) is 10/3