integral of (2x-4)/(x^3+4x^2)

解答

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

\frac{3}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣ + \frac{1}{x} + C

求解步骤

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

乘开

\frac{2 x - 4}{x ^{3} + 4 x ^{2}} : \frac{2 x}{x ^{3} + 4 x ^{2}} - \frac{4}{x ^{3} + 4 x ^{2}}

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

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

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

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

化简

2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣) - (- \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣) : \frac{3}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣ + \frac{1}{x}

= \frac{3}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣ + \frac{1}{x}

解答补常数

= \frac{3}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣ + \frac{1}{x} + C

\frac{d}{d x} (cos (sin (2 x)))

\frac{d}{d x} (a cos (7 x) + b sin (7 x))

n = 1 \sum \infty \frac{2 ^{- n}}{5 ^{- n}}

\frac{d}{d x} (5 x + 3)

l a pl a ce t r an s f or m t^{2} e^{t}