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

The integral from 1 to 3 of ((x^3-1))/(x^4) is ln(3)-26/81

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

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

ln (3) - \frac{26}{81}

Decimal

0.77762 \dots

Solution steps

\int_{1}^{3} \frac{x ^{3} - 1}{x ^{4}} d x

Expand

\frac{x ^{3} - 1}{x ^{4}} : \frac{1}{x} - \frac{1}{x ^{4}}

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

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

\int_{1}^{3} \frac{1}{x ^{4}} d x = \frac{26}{81}

= ln (3) - \frac{26}{81}

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What is the integral from 1 to 3 of ((x^3-1))/(x^4) ?
The integral from 1 to 3 of ((x^3-1))/(x^4) is ln(3)-26/81