What is the integral from 1 to 3 of ((3x^2-7)/(x^3)) ?

The integral from 1 to 3 of ((3x^2-7)/(x^3)) is 3ln(3)-28/9

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integral from 1 to 3 of ((3x^2-7)/(x^3))

\int_{1}^{3} (\frac{3 x ^{2} - 7}{x ^{3}}) d x

3 ln (3) - \frac{28}{9}

Decimal Notation

0.18472 \dots

Solution steps

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

Expand

\frac{3 x ^{2} - 7}{x ^{3}} : \frac{3}{x} - \frac{7}{x ^{3}}

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

\int_{1}^{3} \frac{3}{x} d x = 3 ln (3)

\int_{1}^{3} \frac{7}{x ^{3}} d x = \frac{28}{9}

= 3 ln (3) - \frac{28}{9}

\int t^{3} (7 + t^{4})^{3} d t

x \to 1 lim ((x)^{\frac{1}{x - 1}})

\int_{0}^{4} 3 x d x

\int x^{2} 36 - x^{2} d x

n = 2 \sum \infty n^{2}

What is the integral from 1 to 3 of ((3x^2-7)/(x^3)) ?
The integral from 1 to 3 of ((3x^2-7)/(x^3)) is 3ln(3)-28/9