What is the integral from 1 to 2 of [(inx)/(x^2)] ?

The integral from 1 to 2 of [(inx)/(x^2)] is inln(2)

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integral from 1 to 2 of [(inx)/(x^2)]

\int_{1}^{2} [\frac{in x}{x ^{2}}] d x

in ln (2)

Solution steps

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

Take the constant out:

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

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

Cancel the common factor:

x

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= in [ln ∣ x ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= in ln (2)

What is the integral from 1 to 2 of [(inx)/(x^2)] ?
The integral from 1 to 2 of [(inx)/(x^2)] is inln(2)