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

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

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

\int_{1}^{3} \frac{x}{4 x ^{2}} d x

\frac{1}{4} ln (3)

Decimal Notation

0.27465 \dots

Solution steps

\int_{1}^{3} \frac{x}{4 x ^{2}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int_{1}^{3} \frac{x}{x ^{2}} d x

Cancel the common factor:

x

= \frac{1}{4} \cdot \int_{1}^{3} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= \frac{1}{4} [ln ∣ x ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= \frac{1}{4} ln (3)

\int_{0}^{1} \frac{5}{7 + 7 x ^{2}} d x

\int_{0}^{4} (1 - x) d x

\int_{- 1}^{1} 1 - x d x

\int_{- 1}^{1} (x^{4} - x^{5}) d x

\int_{0}^{\frac{π}{2}} sec^{4} (\frac{x}{2}) d x

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