What is the integral from 0 to 1 of x/(x^2-9) ?

The integral from 0 to 1 of x/(x^2-9) is 3/2 ln(2)-ln(3)

Popular >

integral from 0 to 1 of x/(x^2-9)

\int_{0}^{1} \frac{x}{x ^{2} - 9} d x

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

Decimal

- 0.05889 \dots

Solution steps

\int_{0}^{1} \frac{x}{x ^{2} - 9} d x

Apply u-substitution

: \int_{- 9}^{- 8} \frac{1}{2 u} d u

= \int_{- 9}^{- 8} \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_{- 9}^{- 8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{- 9}^{- 8}

Compute the boundaries:

3 ln (2) - 2 ln (3)

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

Simplify

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

\int_{0}^{2} 6 e^{2 x} d x

\int_{- 4}^{4} (x^{2} + x - 6) d x

\int_{4}^{9} x x - 3 d x

\int_{2}^{4} 16 - p^{2} d p

\int_{0}^{1} y^{2} \cdot 3 (1 - y)^{2} d y

What is the integral from 0 to 1 of x/(x^2-9) ?
The integral from 0 to 1 of x/(x^2-9) is 3/2 ln(2)-ln(3)