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

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

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

\int_{1}^{4} (\frac{1}{x})^{2} d x

\frac{3}{4}

Solution steps

\int_{1}^{4} (\frac{1}{x})^{2} d x

Simplify

(\frac{1}{x})^{2} : \frac{1}{x ^{2}}

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

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{4}

= [- \frac{1}{x}]_{1}^{4}

Compute the boundaries:

\frac{3}{4}

= \frac{3}{4}

y x + y^{2} = 2 x, (4, 2)

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

x (4 x + 3)^{\frac{1}{3}}

e^{x} - 7 x^{6}

cot (6 - u)

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