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

The integral from-2 to 1 of x/(x^2+1) is -1/2 (ln(5)-ln(2))

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

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

- \frac{1}{2} (ln (5) - ln (2))

Decimal

- 0.45814 \dots

Solution steps

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

Apply u-substitution

= \int_{5}^{2} \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{2}^{5} \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}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} [ln ∣ u ∣]_{2}^{5}

Compute the boundaries:

ln (5) - ln (2)

= - \frac{1}{2} (ln (5) - ln (2))

What is the integral from-2 to 1 of x/(x^2+1) ?
The integral from-2 to 1 of x/(x^2+1) is -1/2 (ln(5)-ln(2))