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

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

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

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

\frac{1}{2} (ln (3) - 3 ln (2))

Decimal Notation

- 0.49041 \dots

Solution steps

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

Apply u-substitution

: \int_{- 3}^{- 8} - \frac{1}{2 u} d u

= \int_{- 3}^{- 8} - \frac{1}{2 u} d u

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

= - \int_{- 8}^{- 3} - \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_{- 8}^{- 3} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{2} [ln ∣ u ∣]_{- 8}^{- 3})

Simplify

= \frac{1}{2} [ln ∣ u ∣]_{- 8}^{- 3}

Compute the boundaries:

ln (3) - 3 ln (2)

= \frac{1}{2} (ln (3) - 3 ln (2))

\int_{2}^{6} x d x

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

\int_{0}^{π} sin (x) \cdot sin (n x) d x

\int_{0}^{1} - x + 1 d x

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