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

The integral from 1/2 to 1 of (y+4)/(y^2+y) is 4ln(2)-ln(64/27)

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

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

4 ln (2) - ln (\frac{64}{27})

Decimal Notation

1.90954 \dots

Solution steps

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

Take the partial fraction of

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

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

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

\int_{\frac{1}{2}}^{1} \frac{3}{y + 1} d y = ln (\frac{64}{27})

= 4 ln (2) - ln (\frac{64}{27})

y = 0.7 (1.02)^{x}

f (x) = 1 - 4 x^{2}

\frac{d y}{d x} = y (0.1 - 1 0^{- 7} y)

\frac{d}{d x} (ln ∣ x - 1 ∣ - ln ∣ x - 3 ∣)

\int arccsc (\frac{t}{2}) d t

What is the integral from 1/2 to 1 of (y+4)/(y^2+y) ?
The integral from 1/2 to 1 of (y+4)/(y^2+y) is 4ln(2)-ln(64/27)