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

The integral from 1 to 3 of (4x^4+1)/(4x^2) is 53/6

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

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

\frac{53}{6}

Decimal Notation

8.83333 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

= \frac{1}{4} \cdot \int_{1}^{3} 4 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

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

\int_{1}^{3} 4 x^{2} d x = \frac{104}{3}

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

= \frac{1}{4} (\frac{104}{3} + \frac{2}{3})

Simplify

\frac{1}{4} (\frac{104}{3} + \frac{2}{3}) : \frac{53}{6}

= \frac{53}{6}

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What is the integral from 1 to 3 of (4x^4+1)/(4x^2) ?
The integral from 1 to 3 of (4x^4+1)/(4x^2) is 53/6