integral of (3x^3+9x^2+2x+6)/(x^4+x^2)

解答

\int \frac{3 x ^{3} + 9 x ^{2} + 2 x + 6}{x ^{4} + x ^{2}} d x

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

求解步骤

\int \frac{3 x ^{3} + 9 x ^{2} + 2 x + 6}{x ^{4} + x ^{2}} d x

将

\frac{3 x ^{3} + 9 x ^{2} + 2 x + 6}{x ^{4} + x ^{2}}

用部份分式展开:

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{6}{x ^{2}} d x = - \frac{6}{x}

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

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

解答补常数

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

t a y l or x \cdot sec (x)

\int \frac{1}{x 16 x ^{2} - 9} d x

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