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

The integral from 0 to 2 of x/(x^2+3) is 1/2 (ln(7)-ln(3))

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integral from 0 to 2 of x/(x^2+3)

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

\frac{1}{2} (ln (7) - ln (3))

Decimal Notation

0.42364 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{7} \frac{1}{2 u} d u

= \int_{3}^{7} \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_{3}^{7} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{3}^{7}

Compute the boundaries:

ln (7) - ln (3)

= \frac{1}{2} (ln (7) - ln (3))

\int_{- 1}^{1} x + y + 1 d x

\int_{1}^{x^{3}} e^{t^{2}} d t

\int_{0}^{e - 1} x ln (x + 1) d x

\int_{1}^{2} p^{2} ln (p) d p

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

What is the integral from 0 to 2 of x/(x^2+3) ?
The integral from 0 to 2 of x/(x^2+3) is 1/2 (ln(7)-ln(3))