integral of 6/((x^2+1)(x^2+4))

解答

\int \frac{6}{( x ^{2} + 1 ) ( x ^{2} + 4 )} d x

2 arctan (x) - arctan (\frac{x}{2}) + C

求解步骤

\int \frac{6}{( x ^{2} + 1 ) ( x ^{2} + 4 )} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{1}{( x ^{2} + 1 ) ( x ^{2} + 4 )} d x

将

\frac{1}{( x ^{2} + 1 ) ( x ^{2} + 4 )}

用部份分式展开:

\frac{1}{3 ( x ^{2} + 1 )} - \frac{1}{3 ( x ^{2} + 4 )}

= 6 \cdot \int \frac{1}{3 ( x ^{2} + 1 )} - \frac{1}{3 ( x ^{2} + 4 )} 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

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

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

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

= 6 (\frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2}))

化简

6 (\frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2})) : 2 arctan (x) - arctan (\frac{x}{2})

= 2 arctan (x) - arctan (\frac{x}{2})

解答补常数

= 2 arctan (x) - arctan (\frac{x}{2}) + C

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