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

The integral from 1 to 3 of (4(x^2+1))/(x^2) is 32/3

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

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

\frac{32}{3}

Solution steps

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

Take the constant out:

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

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

\frac{x ^{2} + 1}{x ^{2}} = \frac{1}{x ^{2}} + 1

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{x ^{2}} = x^{- 2}

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

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

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

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

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

Simplify

4 (\frac{2}{3} + 2) : \frac{32}{3}

= \frac{32}{3}

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What is the integral from 1 to 3 of (4(x^2+1))/(x^2) ?
The integral from 1 to 3 of (4(x^2+1))/(x^2) is 32/3