integral of (2x+6)/(2x^2+3x+1)

解答

\int \frac{2 x + 6}{2 x ^{2} + 3 x + 1} d x

\frac{1}{2} ln 16 x^{2} + 24 x + 8 - \frac{9}{2} ln ∣ 4 x + 4 ∣ + \frac{9}{2} ln ∣ 4 x + 2 ∣ + C

求解步骤

\int \frac{2 x + 6}{2 x ^{2} + 3 x + 1} d x

乘开

\frac{2 x + 6}{2 x ^{2} + 3 x + 1} : \frac{2 x}{2 x ^{2} + 3 x + 1} + \frac{6}{2 x ^{2} + 3 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{2 x}{2 x ^{2} + 3 x + 1} d x + \int \frac{6}{2 x ^{2} + 3 x + 1} d x

\int \frac{2 x}{2 x ^{2} + 3 x + 1} d x = 4 (\frac{1}{8} ln 16 x^{2} + 24 x + 8 + \frac{3}{8} (ln ∣ 4 x + 4 ∣ - ln ∣ 4 x + 2 ∣))

\int \frac{6}{2 x ^{2} + 3 x + 1} d x = - 12 (\frac{1}{2} ln ∣ 4 x + 4 ∣ - \frac{1}{2} ln ∣ 4 x + 2 ∣)

= 4 (\frac{1}{8} ln 16 x^{2} + 24 x + 8 + \frac{3}{8} (ln ∣ 4 x + 4 ∣ - ln ∣ 4 x + 2 ∣)) - 12 (\frac{1}{2} ln ∣ 4 x + 4 ∣ - \frac{1}{2} ln ∣ 4 x + 2 ∣)

化简

4 (\frac{1}{8} ln 16 x^{2} + 24 x + 8 + \frac{3}{8} (ln ∣ 4 x + 4 ∣ - ln ∣ 4 x + 2 ∣)) - 12 (\frac{1}{2} ln ∣ 4 x + 4 ∣ - \frac{1}{2} ln ∣ 4 x + 2 ∣) : \frac{1}{2} ln 16 x^{2} + 24 x + 8 - \frac{9}{2} ln ∣ 4 x + 4 ∣ + \frac{9}{2} ln ∣ 4 x + 2 ∣

= \frac{1}{2} ln 16 x^{2} + 24 x + 8 - \frac{9}{2} ln ∣ 4 x + 4 ∣ + \frac{9}{2} ln ∣ 4 x + 2 ∣

解答补常数

= \frac{1}{2} ln 16 x^{2} + 24 x + 8 - \frac{9}{2} ln ∣ 4 x + 4 ∣ + \frac{9}{2} ln ∣ 4 x + 2 ∣ + C

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

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