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}

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}

f (x) = x^{3} e^{4 x}

y^{^{'}} + x (y^{2} + y) = 0, y (2) = 1

\frac{d}{d x} (\frac{( x ^{3} - 4 x ) ^{8}}{8})

x^{2} + 4 x y + x (\frac{d y}{d x}) = 0

\frac{\partial}{\partial x} (y ln (5 x + y))

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