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

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

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

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

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

Decimal

0.45814 \dots

Solution steps

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

Apply u-substitution

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