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

解答

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

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

十进制

1.90954 \dots

求解步骤

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

将

\frac{x + 4}{x ^{2} + x}

用部份分式展开:

\frac{4}{x} - \frac{3}{x + 1}

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

使用积分加法定则:

\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)