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

The integral from 2 to 3 of x/(x^2-3) is 1/2 ln(6)

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

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

\frac{1}{2} ln (6)

Decimal

0.89587 \dots

Solution steps

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

Apply u-substitution

= \int_{1}^{6} \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_{1}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{6}

Compute the boundaries:

ln (6)

= \frac{1}{2} ln (6)

d er i v a t i v e e^{a x + b}

\int sec^{2} (2 x) tan^{5} (2 x) d x

d er i v a t i v e y = 4 x^{2} + 3 x

a re a 1 - x^{2}, x^{2} - 3, x + 3

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

What is the integral from 2 to 3 of x/(x^2-3) ?
The integral from 2 to 3 of x/(x^2-3) is 1/2 ln(6)