What is the integral from 4 to 8 of x/(x^2-1) ?

The integral from 4 to 8 of x/(x^2-1) is 1/2 ln(21/5)

Popular >

integral from 4 to 8 of x/(x^2-1)

\int_{4}^{8} \frac{x}{x ^{2} - 1} d x

\frac{1}{2} ln (\frac{21}{5})

Decimal Notation

0.71754 \dots

Solution steps

\int_{4}^{8} \frac{x}{x ^{2} - 1} d x

Apply u-substitution

: \int_{15}^{63} \frac{1}{2 u} d u

= \int_{15}^{63} \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_{15}^{63} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{15}^{63}

Compute the boundaries:

ln (63) - ln (15)

= \frac{1}{2} (ln (63) - ln (15))

Simplify

= \frac{1}{2} ln (\frac{21}{5})

\frac{d}{d x} (3 x e^{x})

y = e^{8 x} cos (11 x)

f (x) = 5 (1 - x^{2})^{9}

\int \frac{1}{( 5 x - 3 ) ( 3 - 4 x )} d x

\frac{\partial}{\partial x} (\frac{14 x}{7 x ^{2} + y ^{2} + 6})

What is the integral from 4 to 8 of x/(x^2-1) ?
The integral from 4 to 8 of x/(x^2-1) is 1/2 ln(21/5)