What is the integral from 1 to 3 of (x^3+4x)/(x^5) ?

The integral from 1 to 3 of (x^3+4x)/(x^5) is 158/81

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integral from 1 to 3 of (x^3+4x)/(x^5)

\int_{1}^{3} \frac{x ^{3} + 4 x}{x ^{5}} d x

\frac{158}{81}

Decimal

1.95061 \dots

Solution steps

\int_{1}^{3} \frac{x ^{3} + 4 x}{x ^{5}} d x

Simplify

\frac{x ^{3} + 4 x}{x ^{5}} : \frac{x ^{2} + 4}{x ^{4}}

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

Expand

\frac{x ^{2} + 4}{x ^{4}} : \frac{1}{x ^{2}} + \frac{4}{x ^{4}}

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

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

\int_{1}^{3} \frac{4}{x ^{4}} d x = \frac{104}{81}

= \frac{2}{3} + \frac{104}{81}

Simplify

\frac{2}{3} + \frac{104}{81} : \frac{158}{81}

= \frac{158}{81}

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What is the integral from 1 to 3 of (x^3+4x)/(x^5) ?
The integral from 1 to 3 of (x^3+4x)/(x^5) is 158/81