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

The integral from 1 to 4 of (x^2+1/(4x^2)) is 339/16

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

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

\frac{339}{16}

Decimal Notation

21.1875

Solution steps

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

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} \frac{1}{4 x ^{2}} d x = \frac{3}{16}

= 21 + \frac{3}{16}

Simplify

21 + \frac{3}{16} : \frac{339}{16}

= \frac{339}{16}

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What is the integral from 1 to 4 of (x^2+1/(4x^2)) ?
The integral from 1 to 4 of (x^2+1/(4x^2)) is 339/16