integral of ((8x-3))/(x^2+x+1)

解答

\int \frac{( 8 x - 3 )}{x ^{2} + x + 1} d x

16 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) - 23 arctan (\frac{1}{3} (2 x + 1)) + C

求解步骤

\int \frac{8 x - 3}{x ^{2} + x + 1} d x

乘开

\frac{8 x - 3}{x ^{2} + x + 1} : \frac{8 x}{x ^{2} + x + 1} - \frac{3}{x ^{2} + x + 1}

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

\int \frac{8 x}{x ^{2} + x + 1} d x = 16 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3}))

\int \frac{3}{x ^{2} + x + 1} d x = 23 arctan (\frac{1}{3} (2 x + 1))

= 16 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) - 23 arctan (\frac{1}{3} (2 x + 1))

解答补常数

= 16 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) - 23 arctan (\frac{1}{3} (2 x + 1)) + C

n = 1 \sum \infty 2^{n} - 3

\frac{d}{d x} (8^{x^{3}})

\frac{d y}{d t} = 4 y + y^{3}

\int e^{n x} sin (n x) d x

x \to 0 lim (x sin (\frac{6}{x}))