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

The integral from-1 to-4 of (1-x^4)/(x^2) is 81/4

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

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

\frac{81}{4}

Decimal Notation

20.25

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Expand

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

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

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

\int_{- 4}^{- 1} x^{2} d x = 21

= - (\frac{3}{4} - 21)

Simplify

- (\frac{3}{4} - 21) : \frac{81}{4}

= \frac{81}{4}

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What is the integral from-1 to-4 of (1-x^4)/(x^2) ?
The integral from-1 to-4 of (1-x^4)/(x^2) is 81/4