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

The integral from 0 to 4 of x/(2x^2+2) is 1/4 ln(17)

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

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

\frac{1}{4} ln (17)

Decimal

0.70830 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{34} \frac{1}{4 u} d u

= \int_{2}^{34} \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_{2}^{34} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} [ln ∣ u ∣]_{2}^{34}

Compute the boundaries:

ln (34) - ln (2)

= \frac{1}{4} (ln (34) - ln (2))

Simplify

= \frac{1}{4} ln (17)

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What is the integral from 0 to 4 of x/(2x^2+2) ?
The integral from 0 to 4 of x/(2x^2+2) is 1/4 ln(17)