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

The integral from 1 to 3 of ((3x+1)^2)/(x^3) is 9ln(3)+40/9

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

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

9 ln (3) + \frac{40}{9}

Decimal

14.33195 \dots

Solution steps

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

Expand

\frac{( 3 x + 1 ) ^{2}}{x ^{3}} : \frac{9}{x} + \frac{6}{x ^{2}} + \frac{1}{x ^{3}}

= \int_{1}^{3} \frac{9}{x} + \frac{6}{x ^{2}} + \frac{1}{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{9}{x} d x + \int_{1}^{3} \frac{6}{x ^{2}} d x + \int_{1}^{3} \frac{1}{x ^{3}} d x

\int_{1}^{3} \frac{9}{x} d x = 9 ln (3)

\int_{1}^{3} \frac{6}{x ^{2}} d x = 4

\int_{1}^{3} \frac{1}{x ^{3}} d x = \frac{4}{9}

= 9 ln (3) + 4 + \frac{4}{9}

Combine the fractions

4 + \frac{4}{9} : \frac{40}{9}

= 9 ln (3) + \frac{40}{9}

What is the integral from 1 to 3 of ((3x+1)^2)/(x^3) ?
The integral from 1 to 3 of ((3x+1)^2)/(x^3) is 9ln(3)+40/9