What is the integral from 1 to 3 of 9/(x^2) ?

The integral from 1 to 3 of 9/(x^2) is 6

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integral from 1 to 3 of 9/(x^2)

\int_{1}^{3} \frac{9}{x ^{2}} d x

6

Solution steps

\int_{1}^{3} \frac{9}{x ^{2}} d x

Take the constant out:

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

= 9 \cdot \int_{1}^{3} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{3}

= 9 [- \frac{1}{x}]_{1}^{3}

Compute the boundaries:

\frac{2}{3}

= 9 \cdot \frac{2}{3}

Simplify

= 6

\int e^{- 4 x} sin (5 x) d x

f (x) = \frac{110}{x ^{12}}

f (x) = 7 x^{4} arctan (4 x^{4})

\int \frac{7}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

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

What is the integral from 1 to 3 of 9/(x^2) ?
The integral from 1 to 3 of 9/(x^2) is 6