What is the integral from 1 to x of (10)/(9x^2) ?

The integral from 1 to x of (10)/(9x^2) is 10/9 (-1/x+1)

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integral from 1 to x of (10)/(9x^2)

\int_{1}^{x} \frac{10}{9 x ^{2}} d x

\frac{10}{9} (- \frac{1}{x} + 1)

Solution steps

\int_{1}^{x} \frac{10}{9 x ^{2}} d x

Take the constant out:

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

= \frac{10}{9} \cdot \int_{1}^{x} \frac{1}{x ^{2}} d x

Apply the Power Rule

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

= \frac{10}{9} [- \frac{1}{x}]_{1}^{x}

Compute the boundaries:

- \frac{1}{x} + 1

= \frac{10}{9} (- \frac{1}{x} + 1)

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What is the integral from 1 to x of (10)/(9x^2) ?
The integral from 1 to x of (10)/(9x^2) is 10/9 (-1/x+1)