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

The integral from 0 to 2 of x/(2x^2+4) is 1/4 ln(3)

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

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

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

Decimal Notation

0.27465 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{12} \frac{1}{4 u} d u

= \int_{4}^{12} \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{4}^{12} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} [ln ∣ u ∣]_{4}^{12}

Compute the boundaries:

ln (12) - 2 ln (2)

= \frac{1}{4} (ln (12) - 2 ln (2))

Simplify

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

\int_{0}^{1} \frac{1}{( x ^{2} + 2 ) ^{2}} d x

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\int_{0}^{4} 8 x^{2} d x

\int_{1}^{7} (t^{2} - 2 t - 24) d t

\int_{0}^{4} x^{2} + 2 x - 3 d x

What is the integral from 0 to 2 of x/(2x^2+4) ?
The integral from 0 to 2 of x/(2x^2+4) is 1/4 ln(3)