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

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

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

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

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

Decimal

0.34657 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{8} \frac{1}{2 u} d u

= \int_{4}^{8} \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_{4}^{8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{4}^{8}

Compute the boundaries:

ln (2)

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

x \to - 3 - lim (\frac{1}{x ^{2} - 9})

\int \frac{1}{4 x ( 6 + x )} d x

\frac{2000 t}{4 t + 75}

\frac{\partial}{\partial x} (x + 2)

\int \frac{15 e ^{5 z}}{e ^{5 z} + 2} d z

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