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

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

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

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

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

Decimal

1.90954 \dots

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\int_{0}^{1} \int 1 d x d y

x \to 4 lim (x + 2)

\int_{1}^{8} π (x - 1) d x

t an g e n t f (x) = 3 x^{2} + 8 x + 4, at x = - 1

x \to 0 lim (x^{3} + 2 x^{2} - x - 2)

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