What is the integral from 1/2 to 1 of (y+2)/(y^2+y) ?

The integral from 1/2 to 1 of (y+2)/(y^2+y) is ln(2)+ln(3/2)

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integral from 1/2 to 1 of (y+2)/(y^2+y)

\int_{\frac{1}{2}}^{1} \frac{y + 2}{y ^{2} + y} d y

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

Decimal

1.09861 \dots

Solution steps

\int_{\frac{1}{2}}^{1} \frac{y + 2}{y ^{2} + y} d y

Take the partial fraction of

\frac{y + 2}{y ^{2} + y} : \frac{2}{y} - \frac{1}{y + 1}

= \int_{\frac{1}{2}}^{1} \frac{2}{y} - \frac{1}{y + 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_{\frac{1}{2}}^{1} \frac{2}{y} d y - \int_{\frac{1}{2}}^{1} \frac{1}{y + 1} d y

\int_{\frac{1}{2}}^{1} \frac{2}{y} d y = 2 ln (2)

\int_{\frac{1}{2}}^{1} \frac{1}{y + 1} d y = ln (2) - ln (\frac{3}{2})

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

Simplify

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

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