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

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

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

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

- 1

Solution steps

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

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

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

= \frac{1}{2} - \frac{3}{2}

Simplify

\frac{1}{2} - \frac{3}{2} : - 1

= - 1

\frac{d}{d t} (f (t))

\frac{d}{d x} (\frac{5 x ^{2} - 3 x ^{5}}{- x})

\int_{- \infty}^{\infty} x d x

\int e^{- 2 x} (x^{2} - 3 x + 4) d x

\frac{s}{s ^{2} - 2 s + 5}

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