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

The integral from 1 to 3 of (x^4+1)/(x^2) is 28/3

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

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

\frac{28}{3}

Decimal Notation

9.33333 \dots

Solution steps

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

Expand

\frac{x ^{4} + 1}{x ^{2}} : x^{2} + \frac{1}{x ^{2}}

= \int_{1}^{3} x^{2} + \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

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

\int_{1}^{3} x^{2} d x = \frac{26}{3}

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

= \frac{26}{3} + \frac{2}{3}

Simplify

\frac{26}{3} + \frac{2}{3} : \frac{28}{3}

= \frac{28}{3}

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\int_{2}^{3} \frac{1}{x + 1} d x

What is the integral from 1 to 3 of (x^4+1)/(x^2) ?
The integral from 1 to 3 of (x^4+1)/(x^2) is 28/3