What is the integral from 1/2 to 1 of (3y+4)/(y^2+y) ?

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

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integral from 1/2 to 1 of (3y+4)/(y^2+y)

\int_{\frac{1}{2}}^{1} \frac{3 y + 4}{y ^{2} + y} d y

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

Decimal Notation

2.48490 \dots

Solution steps

\int_{\frac{1}{2}}^{1} \frac{3 y + 4}{y ^{2} + y} d y

Take the partial fraction of

\frac{3 y + 4}{y ^{2} + y} : \frac{4}{y} - \frac{1}{y + 1}

= \int_{\frac{1}{2}}^{1} \frac{4}{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{4}{y} d y - \int_{\frac{1}{2}}^{1} \frac{1}{y + 1} d y

\int_{\frac{1}{2}}^{1} \frac{4}{y} d y = 4 ln (2)

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

= 4 ln (2) - (ln (2) - ln (\frac{3}{2}))

Simplify

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

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