What is the integral from 1/2 to 1 of (4y+5)/(y^2+y) ?

The integral from 1/2 to 1 of (4y+5)/(y^2+y) is 4ln(2)+ln(3/2)

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integral from 1/2 to 1 of (4y+5)/(y^2+y)

\int_{\frac{1}{2}}^{1} \frac{4 y + 5}{y ^{2} + y} d y

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

Decimal

3.17805 \dots

Solution steps

\int_{\frac{1}{2}}^{1} \frac{4 y + 5}{y ^{2} + y} d y

Take the partial fraction of

\frac{4 y + 5}{y ^{2} + y} : \frac{5}{y} - \frac{1}{y + 1}

= \int_{\frac{1}{2}}^{1} \frac{5}{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{5}{y} d y - \int_{\frac{1}{2}}^{1} \frac{1}{y + 1} d y

\int_{\frac{1}{2}}^{1} \frac{5}{y} d y = 5 ln (2)

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

= 5 ln (2) - (ln (2) - ln (\frac{3}{2}))

Simplify

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

\int_{- 5}^{5} \frac{1}{x ^{(\frac{2}{3})}} d x

(0, - 3), (2, 1)

\frac{1}{3}

\int (x^{2} + 1)^{2} (2 x) d x

y = x 9 - x^{2}, y = 0, x = 0, x = 3

What is the integral from 1/2 to 1 of (4y+5)/(y^2+y) ?
The integral from 1/2 to 1 of (4y+5)/(y^2+y) is 4ln(2)+ln(3/2)