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

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

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

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

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

Decimal Notation

1.79175 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

y^{^{''}} + 9 y = cos (3 t)

\frac{d}{d x} (ln (x^{6} + x^{3} - 6))

\int x^{2} \cdot 1 + x d x

\frac{\partial}{\partial x} (- e^{8 y} cos (8 x) \cdot 8)

y = tan (π x)

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