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

The integral from-1 to 3 of x/(x^2+1) is 1/2 ln(5)

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integral from-1 to 3 of x/(x^2+1)

\int_{- 1}^{3} \frac{x}{x ^{2} + 1} d x

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

Decimal Notation

0.80471 \dots

Solution steps

\int_{- 1}^{3} \frac{x}{x ^{2} + 1} d x

Apply u-substitution

: \int_{2}^{10} \frac{1}{2 u} d u

= \int_{2}^{10} \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}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{2}^{10}

Compute the boundaries:

ln (10) - ln (2)

= \frac{1}{2} (ln (10) - ln (2))

Simplify

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

- 6 cos (t)

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\int_{1}^{- 1} (12 x + 11)^{2} d x

\frac{d}{d x} (arcsin (x ln (x)))

\int_{0}^{1} \frac{5}{1 + x ^{2}} d x

What is the integral from-1 to 3 of x/(x^2+1) ?
The integral from-1 to 3 of x/(x^2+1) is 1/2 ln(5)