integral of 8/((x+2)(x^2+4))

解答

\int \frac{8}{( x + 2 ) ( x ^{2} + 4 )} d x

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

求解步骤

\int \frac{8}{( x + 2 ) ( x ^{2} + 4 )} d x

提出常数:

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

= 8 \cdot \int \frac{1}{( x + 2 ) ( x ^{2} + 4 )} d x

将

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

用部份分式展开:

\frac{1}{8 ( x + 2 )} + \frac{- x + 2}{8 ( x ^{2} + 4 )}

= 8 \cdot \int \frac{1}{8 ( x + 2 )} + \frac{- x + 2}{8 ( 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

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

\int \frac{1}{8 ( x + 2 )} d x = \frac{1}{8} ln ∣ x + 2 ∣

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

= 8 (\frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} (- \frac{1}{2} ln x^{2} + 4 + arctan (\frac{x}{2})))

化简

8 (\frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} (- \frac{1}{2} ln x^{2} + 4 + arctan (\frac{x}{2}))) : ln ∣ x + 2 ∣ - \frac{1}{2} ln x^{2} + 4 + arctan (\frac{x}{2})

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

解答补常数

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

t an g e n t \frac{- 5 x ^{2}}{2 x + 3}

\frac{d}{d x} (arcsin (\frac{a}{x}))

\int \frac{x ^{3} - 2 x ^{2} + 5 x - 2}{x ^{3} - 2 x ^{2} + x} d x

\frac{d}{d x} (- 2 sin^{2} (x))

n = 1 \sum \infty \frac{( ln ( 3 n ) )}{n}