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

解答

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

\frac{3}{4} ln 4 x^{2} + 4 x + 6 - \frac{1}{2 5} arctan (\frac{1}{5} (2 x + 1)) + C

求解步骤

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

乘开

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

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

\int \frac{3 x}{2 x ^{2} + 2 x + 3} d x = 3 (\frac{1}{4} ln 4 x^{2} + 4 x + 6 - \frac{1}{2 5} arctan (\frac{2 x + 1}{5}))

\int \frac{1}{2 x ^{2} + 2 x + 3} d x = \frac{1}{5} arctan (\frac{2 x + 1}{5})

= 3 (\frac{1}{4} ln 4 x^{2} + 4 x + 6 - \frac{1}{2 5} arctan (\frac{2 x + 1}{5})) + \frac{1}{5} arctan (\frac{2 x + 1}{5})

化简

3 (\frac{1}{4} ln 4 x^{2} + 4 x + 6 - \frac{1}{2 5} arctan (\frac{2 x + 1}{5})) + \frac{1}{5} arctan (\frac{2 x + 1}{5}) : \frac{3}{4} ln 4 x^{2} + 4 x + 6 - \frac{1}{2 5} arctan (\frac{1}{5} (2 x + 1))

= \frac{3}{4} ln 4 x^{2} + 4 x + 6 - \frac{1}{2 5} arctan (\frac{1}{5} (2 x + 1))

解答补常数

= \frac{3}{4} ln 4 x^{2} + 4 x + 6 - \frac{1}{2 5} arctan (\frac{1}{5} (2 x + 1)) + C

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