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

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

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

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

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

Decimal Notation

0.34657 \dots

Solution steps

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

Apply u-substitution

: \int_{5}^{10} \frac{1}{2 u} d u

= \int_{5}^{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_{5}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{5}^{10}

Compute the boundaries:

ln (10) - ln (5)

= \frac{1}{2} (ln (10) - ln (5))

Simplify

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

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

\int - \frac{4}{3}

\frac{\partial}{\partial x} (2 x y + y)

\int_{0}^{1} \frac{x - x}{7} d x

\int 2 ln (x) x d x

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