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

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

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integral from 0 to 1 of x/(x^2+2)

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

\frac{1}{2} (ln (3) - ln (2))

Decimal Notation

0.20273 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{3} \frac{1}{2 u} d u

= \int_{2}^{3} \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_{2}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{2}^{3}

Compute the boundaries:

ln (3) - ln (2)

= \frac{1}{2} (ln (3) - ln (2))

\int_{2}^{6} (8 t^{3} - 6 t^{- 2}) d t

\int_{- 1}^{0} (x^{5}) x^{3} + 1 d x

\int_{0}^{1} \frac{x ^{3}}{2} d x

\int_{- 4}^{2} \frac{1}{16 - x ^{2}} d x

\int_{- 1}^{1} - x^{2} d x

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