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

解答

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

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

求解步骤

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

提出常数:

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

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

将

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

用部份分式展开:

\frac{1}{16 x} - \frac{1}{16 ( x - 4 )} + \frac{1}{4 ( x - 4 ) ^{2}}

= 4 \cdot \int \frac{1}{16 x} - \frac{1}{16 ( x - 4 )} + \frac{1}{4 ( x - 4 ) ^{2}} 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}{16 x} d x - \int \frac{1}{16 ( x - 4 )} d x + \int \frac{1}{4 ( x - 4 ) ^{2}} d x)

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

\int \frac{1}{16 ( x - 4 )} d x = \frac{1}{16} ln ∣ x - 4 ∣

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

= 4 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )})

化简

4 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )}) : \frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x - 4 ∣ - \frac{1}{x - 4}

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

解答补常数

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

x \to 1 + lim (arcsin (x))

\frac{d}{d x} (e^{2 x} (3 x - 2))

e x p an d (x^{3} - 4 x^{2} - 6 x + 1) (x^{2} - 3 x + 1)

\frac{d}{d x} (- sin (2 x) + 1)

\frac{d}{d x} ((5 x - 3) (4 x - \frac{3}{x}))