What is the integral from 1/4 to 1 of (y+6)/(y^2+3y) ?

The integral from 1/4 to 1 of (y+6)/(y^2+3y) is 2ln(2)+ln(13/4)

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integral from 1/4 to 1 of (y+6)/(y^2+3y)

\int_{\frac{1}{4}}^{1} \frac{y + 6}{y ^{2} + 3 y} d y

2 ln (2) + ln (\frac{13}{4})

Decimal Notation

2.56494 \dots

Solution steps

\int_{\frac{1}{4}}^{1} \frac{y + 6}{y ^{2} + 3 y} d y

Take the partial fraction of

\frac{y + 6}{y ^{2} + 3 y} : \frac{2}{y} - \frac{1}{y + 3}

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

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

\int_{\frac{1}{4}}^{1} \frac{1}{y + 3} d y = 2 ln (2) - ln (\frac{13}{4})

= 4 ln (2) - (2 ln (2) - ln (\frac{13}{4}))

Simplify

= 2 ln (2) + ln (\frac{13}{4})

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What is the integral from 1/4 to 1 of (y+6)/(y^2+3y) ?
The integral from 1/4 to 1 of (y+6)/(y^2+3y) is 2ln(2)+ln(13/4)