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

The integral from 1 to 3 of ((x-1)^3)/(x^2) is 3ln(3)-8/3

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

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

3 ln (3) - \frac{8}{3}

Decimal Notation

0.62917 \dots

Solution steps

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

Expand

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

= \int_{1}^{3} x - 3 + \frac{3}{x} - \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

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

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

\int_{1}^{3} 3 d x = 6

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

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

= 4 - 6 + 3 ln (3) - \frac{2}{3}

Simplify

4 - 6 + 3 ln (3) - \frac{2}{3} : 3 ln (3) - \frac{8}{3}

= 3 ln (3) - \frac{8}{3}

\int_{0}^{3} (- x^{2} + 2 x + 3) d x

\int_{t}^{3 t} (\frac{sin ( x )}{x}) d x

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

\int_{π}^{2 π} e^{- s t} d t

\int_{2}^{3} 95 + 972 t^{2} - 18 t d t

What is the integral from 1 to 3 of ((x-1)^3)/(x^2) ?
The integral from 1 to 3 of ((x-1)^3)/(x^2) is 3ln(3)-8/3