What is the integral from 1 to 3 of (y+4)/(y-4) ?

The integral from 1 to 3 of (y+4)/(y-4) is -8ln(3)+2

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integral from 1 to 3 of (y+4)/(y-4)

\int_{1}^{3} \frac{y + 4}{y - 4} d y

- 8 ln (3) + 2

Decimal

- 6.78889 \dots

Solution steps

\int_{1}^{3} \frac{y + 4}{y - 4} d y

\frac{y + 4}{y - 4} = \frac{8}{y - 4} + 1

= \int_{1}^{3} \frac{8}{y - 4} + 1 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{3} \frac{8}{y - 4} d y + \int_{1}^{3} 1 d y

\int_{1}^{3} \frac{8}{y - 4} d y = - 8 ln (3)

\int_{1}^{3} 1 d y = 2

= - 8 ln (3) + 2

What is the integral from 1 to 3 of (y+4)/(y-4) ?
The integral from 1 to 3 of (y+4)/(y-4) is -8ln(3)+2