integral of (4x)/(x^3+x^2-x-1)

解答

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

- ln ∣ x + 1 ∣ - \frac{2}{x + 1} + ln ∣ x - 1 ∣ + C

求解步骤

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

提出常数:

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

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

将

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

用部份分式展开:

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

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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

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

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

化简

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

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

解答补常数

= - ln ∣ x + 1 ∣ - \frac{2}{x + 1} + ln ∣ x - 1 ∣ + C

\frac{d}{d x} (- cos (x))

x \to - 1 lim (- x^{2} + 2 x - 9)

- y^{^{'}} + \frac{4 y}{x} = x y^{2}

in v erse l a pl a ce \frac{s}{s ^{2} + 4 s + 3}

\frac{\partial}{\partial x} (4 x^{0.75} + y)