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

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

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

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

- \frac{1}{x ^{2}} + 1

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

: [- \frac{1}{2 x ^{2}}]_{1}^{x}

= 2 [- \frac{1}{2 x ^{2}}]_{1}^{x}

Compute the boundaries:

- \frac{1}{2 x ^{2}} + \frac{1}{2}

= 2 (- \frac{1}{2 x ^{2}} + \frac{1}{2})

Simplify

= - \frac{1}{x ^{2}} + 1

y = \frac{x - 1}{x + 1}

\frac{d}{d x} (7 x^{- 6} - 5 x^{- 3})

\int cot^{7} (x) csc^{2} (x) d x

y = - 10

\frac{d}{d y} (x^{2 y})

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