What is the integral from 0 to 4 of x/(16+x^2) ?

The integral from 0 to 4 of x/(16+x^2) is 1/2 ln(2)

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integral from 0 to 4 of x/(16+x^2)

\int_{0}^{4} \frac{x}{16 + x ^{2}} d x

\frac{1}{2} ln (2)

Decimal Notation

0.34657 \dots

Solution steps

\int_{0}^{4} \frac{x}{16 + x ^{2}} d x

Apply u-substitution

: \int_{16}^{32} \frac{1}{2 u} d u

= \int_{16}^{32} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{16}^{32} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{16}^{32}

Compute the boundaries:

ln (2)

= \frac{1}{2} ln (2)

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\int_{0}^{\frac{7}{4}} \frac{1}{16 x ^{2} + 49} d x

\int_{0}^{5} 75 x^{4} d x

What is the integral from 0 to 4 of x/(16+x^2) ?
The integral from 0 to 4 of x/(16+x^2) is 1/2 ln(2)