What is the integral from 1 to 3 of (x-4x^2)/(x^3) ?

The integral from 1 to 3 of (x-4x^2)/(x^3) is 2/3-4ln(3)

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integral from 1 to 3 of (x-4x^2)/(x^3)

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

\frac{2}{3} - 4 ln (3)

Decimal

- 3.72778 \dots

Solution steps

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

\frac{x - 4 x ^{2}}{x ^{3}} = (\frac{x}{x ^{3}} - \frac{4 x ^{2}}{x ^{3}})

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

Simplify

\frac{x}{x ^{3}} - \frac{4 x ^{2}}{x ^{3}} : \frac{1}{x ^{2}} - \frac{4}{x}

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

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

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

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

\int_{- π}^{0} cos (n) x d x

\int_{0}^{5} x^{2} + 2 x - 3 d x

\int_{0}^{1} 0.5 e^{x^{2}} d x

\int_{0}^{1} cos (nπ t) d t

\int_{0}^{1} (2 e^{x}) d x

What is the integral from 1 to 3 of (x-4x^2)/(x^3) ?
The integral from 1 to 3 of (x-4x^2)/(x^3) is 2/3-4ln(3)